Pls help me answer these questions⬇⬇⬇

The value of x is calculated with the help of a similar triangle is 24.

A triangle is a three-sided polygon with three angles. The angles of the triangle add up to 180 degrees.

In a triangle, if there is a parallel line with any side of the triangle, then the line divides the other sides in the same ratio.

Then we have

x / 12 = 32 / 16

     x = 24

Then the value of x is 24.

More about the triangle link is given below.

https://brainly.com/question/25813512

#SPJ1

Answ x=59

Step-by-step explanation:

Answer:

D.  [tex](4x^2 + 3)(x + 2)[/tex]

Step-by-step explanation:

Hello!

We can group the first two terms and the last two terms and factor each group.

Now we can combine the like factors:

The answer is Option D.

Answer:

J Trump has to be a great day of school and I have a great

Step-by-step explanation:

you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you for your loss of school and I have a great day of school ii

Answer:

Step-by-step explanation:

Jakub wants to build a new shed.

The area of the shed is 37 m².

The width of the shed is 5 m.

What is the length of the shed?

to find the side, having the area, you need to use the inverse formula Area = W * L

so

L = A: W

L = 37 : 5 = 7.4 m

If the area of the shed is 37 m²,width of the shed is 5 m as a result the length of the shed will be 7.4 m.

It is defined as two-dimensional geometry in which the angle between the adjacent sides is 90 degrees. It is a type of quadrilateral. The area occupied by the rectangle in two-dimensional planner geometry.

The area of a rectangle can be calculated using the following formula:

Rectangle area = length x width

It is given that, the area of the shed is 37 m². The width of the shed is 5 m.

Since the shed is rectangular. The length of width is found as,

Area = length × width

A = l × w

l=A/w

l=37/5

l=7.4 m

Thus, if the area of the shed is 37 m², the width of the shed is 5 m as a result the length of the shed will be 7.4 m.

Learn more about the rectangle here:

https://brainly.com/question/15019502

#SPJ2

Answer:

y≤x²+4x-1.

Step-by-step explanation:

1) to define the equation of the given graph; it is y=x²+4x-1 (its vertex is (-2;-5));

2) to define inequation accorging to the given graph (the area outside of the parabola means ≤).

he value of x and y from the given figure are 49/5 and 49/15

From the given similar triangles, the following expression is true

21/30 = 7/15 = k

Also, x/21 = y/7 = 7/15

Equate

x/21 = 7/15

15x = 7 * 21

5x = 7 * 7

x = 49/5

Similarly

y/7 = 7/15

15y = 49

y = 49/15

Hence the value of x and y from the given figure are 49/5 and 49/15

Learn more on similar triangle here; https://brainly.com/question/14285697

#SPJ1

Answer:

c6

Step-by-step explanation:

justjustcancacanjustadda0totheproduct of5cxwewtogetthesameresultasx1totoandtimes

3) Number: 640,700

   the value of 4 in this number is 40,000

Number: 64,070

the value of 4 in this number is 4,000

So, times greater: 40,000 ÷ 4,000 = 10 times greater.

4)

The multiplication, 5 × 12 = 60. So, 5 × 120 = 600.

There is an extra zero at the very last of the result as '120' was multiplied with 5 instead of '12' which does not have zero at last which changes the result.

Similar cases with: 5 × 10 = 50. So, 5 × 100 = 500.

The stock worth now, to the nearest cent, is $58.14.

The current worth of stock = initial price of stock+increase in the price of stocks

If you bought stock last year for a price of $66, and it has gone down 13.9% since then,

Let X be the worth of stock;

66 ×  100 = (13.5% + 100%) × X

6600 = 113.5X

X = 6600/113.5

X = 58.14458

X = $58.14

Hence, The stock worth now, to the nearest cent is $58.14.

Learn more about the stock;

brainly.com/question/16553868

#SPJ1

The sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

For a sector that has a central angle of θ, and a radius of r;

The sector area, and the arc length are:

[tex]A = \frac{\theta}{360} * \pi r^2[/tex] --- sector area

[tex]L = \frac{\theta}{360} * 2\pi r[/tex] ---- arc length

Here, the given parameters are:

Central angle, θ = 160

Radius, r = 5 inches

The sector area is

[tex]A = \frac{\theta}{360} * \pi r^2[/tex]

So, we have:

[tex]A = \frac{160}{360} * \frac{22}{7} * 5^2[/tex]

Evaluate

A = 34.92

The arc length is:

[tex]L = \frac{\theta}{360} * 2\pi r[/tex]

So, we have:

[tex]L = \frac{160}{360} * 2 * \frac{22}{7} * 5[/tex]

L = 13.97

Hence, the sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

Read more about sector area and arc length at:

https://brainly.com/question/2005046

#SPJ1

The zeroes of the quadratic function f(x) = 6x² + 12x – 7 will be given below. Then the correct option is A.

The quadratic function is given as ax² + bx + c = f(x).

For finding the zeroes of the polynomial we have to find the roots of the polynomial , The value of f(x) = 0

The given function is f(x) = 6x² + 12x – 7

Then the zeroes of the quadratic function will be

[tex]roots = \dfrac {-b \pm \sqrt{b^2 -4ac}}{2a}\\\\[/tex]

Here a =6 , b = 12 ,c = -7

Therefore the zeroes are

[tex]\rm roots = \dfrac {-12 \pm \sqrt{12^2 -4 * 6 * (-7)}}{2 *6}\\\\[/tex]

[tex]\rm root_1 = -1 + \sqrt{\dfrac {13}{6}}\\\\\rm root_2 = -1 - \sqrt{\dfrac {13}{6}}[/tex]

Therefore the correct option is A.

To know more about Quadratic Function

brainly.com/question/17376136

#SPJ1

Answer:

the answer is c. the row is chili, the column is no cheese

[tex]z_{2}-3i=(-4+i)-3i=-4-2i\\\\z_{1}+2=(3-4i)+2=5-4i\\\\(z_{1}+2)-(z_{2}-3i)=(5-4i)-(-4-2i)=9-4i+4+2i=9-2i[/tex]

which corresponds with point C

The expression which is the result of subtracting (z2-3i) from (z1+2) is 9-2i.

Given On a coordinate plane, the y-axis is labeled imaginary and the x-axis is labeled real. Point A is (1, -2), B is (1, -6), C is (9,- 2), D is (9, - 6), z 1 is (3, - 4), and z 2 is (- 4, 1).

Coordinate plane is a two dimensional plane formed by the intersection of two number lines. One is horizontal line and one is vertical. On horizontal line x values are denoted and on vertical line y values are written. There can be three dimensional plane also.

z2-3i=(-4+i)-3i=-4-2i

z1+2=(3-4i)+2=5-4i

(z1+2)-(z2-3i)=(5-4i)-(-4-2i)

=9-4i+4+2i

=9-2i

Hence for the expression (z1+2)-(z2-3i)=9-2i

Learn more about coordinate plane at  https://brainly.com/question/13854649

#SPJ10

The graph of the function is shown in the attached image.

Answer:

7

4

Step-by-step explanation:

The actual values are shown on the given graph as blue points.

The line of regression is shown on the given graph as the red line.

From inspection of the graph, in the year 2000 the actual rainfall was 43 cm, shown by point (2000, 43).  It appears that the regression line is at y = 50 when x is the year 2000.

⇒ Difference = 50 - 43 = 7 cm

In 2000, the actual rainfall was 7 centimeters below what the model predicts.

From inspection of the graph, in the year 2003 the actual rainfall was 44 cm,  shown by point (2003, 40).  It appears that the regression line is at y = 40 when x is the year 2003.

⇒ Difference = 44 - 40 = 4 cm

In 2003, the actual rainfall was 4 centimeters above what the model predicts.

The school bus will take 5.34 seconds to travel 95.1 m if the bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s².

Any equation of the form [tex]\rm ax^2+bx+c=0[/tex] where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

[tex]\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}[/tex]

We have:

A school bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s².

From the second equation of motion:

[tex]\rm s = u + \dfrac{1}{2}at^2[/tex]

We have given:

s = 95.1 m

u = 26.8 m/s

a = 4.73 m/s²

[tex]\rm 95.1 = 26.8 + \dfrac{1}{2}(4.73)t^2[/tex]

The above equation is a quadratic equation:

After simplification:

[tex]\rm \dfrac{4.73}{2}t^2=68.3[/tex]

[tex]\rm t^2=28.879[/tex]

t = 5.373 seconds ≈ 5.34 seconds

Thus, the school bus will take 5.34 seconds to travel 95.1 m if the bus is moving 26.8 m/s on flat ground when it begins to accelerate at 4.73 m/s².

Learn more about quadratic equations here:

brainly.com/question/2263981

#SPJ1

Answer:

See proof below

Step-by-step explanation:

Important points

Onto

For a function with a given co-domain to be "onto," every element of the co-domain must be an element of the range.

However, the co-domain here is suggested to be [tex]\mathbb R[/tex], whereas the range of f is not [tex]\mathbb R[/tex] (proof below).

Proof (contradiction)

Suppose that f is onto [tex]\mathbb R[/tex].

Consider the output 7 (a specific element of [tex]\mathbb R[/tex]).

Since f is onto [tex]\mathbb R[/tex], there must exist some input from the domain [tex]\mathbb R[/tex], "p", such that f(p) = 7.

Substitute and solve to find values for "p".

[tex]f(x)=-3x^2+4\\f(p)=-3(p)^2+4\\7=-3p^2+4\\3=-3p^2\\-1=p^2[/tex]

Next, apply the square root property:

[tex]\pm \sqrt{-1} =\sqrt{p^2}[/tex]

By definition, [tex]\sqrt{-1} =i[/tex], so

[tex]i=p \text{ or } -i =p[/tex]

By the Fundamental Theorem of Algebra, any polynomial of degree n with complex coefficients, has exactly n complex roots.  Since the degree of f is 2, there are exactly 2 roots, and we've found them both, so we've found all of them.

However, neither [tex]i[/tex]  nor [tex]-i[/tex] are in [tex]\mathbb R[/tex], so there are zero values of p in [tex]\mathbb R[/tex] for which f(p)= 7, which is a contradiction.

Therefore, the contradiction supposition must be false, proving that f is not onto [tex]\mathbb R[/tex]

Answer: 2.7545

The question was a little off, but I can understand enough

Step-by-step explanation:

Grade       Grade Point(x)       Course Credit(y)        Product(xy)

A                     4                            3                                12

B                      3                           4                                12

B                     3                            4                                12

B                      3                          4                                 12

C                     2                           2                                 4

D                      1                           3                                 3

Total                                              20                             55

This means, the GPA of the student is :

[tex]GPA=\frac{55}{20}[/tex]

[tex]2.75[/tex]

Answer:

bro nobody can figure this out you need to word the entire question with numbers and all. and proper English pls

Answer:

19 small frames and 3 large frames

Step-by-step explanation:

follows the steps in the attached

Answer:

The balance after 37 years is $3869.99

Step-by-step explanation:

Given:

The Amount (A),

A = P (1 + [tex]\frac{R}{100n}[/tex][tex])^{nt}[/tex]        if compound n times per year

Amount (A) at 37 years,

A = ($2,520) (1 + [tex]\frac{1.16}{100(12)}[/tex][tex])^{12 x 37}[/tex]

A = ($2,520) (1 + [tex]\frac{1.16}{1,200}[/tex][tex])^{444}[/tex]

A = ($2,520) (1.00096667 [tex])^{444}[/tex]

A = ($2,520) (1.5357)

A = $3869.9944

Learn more about Compounds Interest here: https://brainly.com/question/12325365

Answer:

1.  "a"  [tex]u'=6x[/tex]

2.  "d"  [tex]v'=15x^2[/tex]

3.  "b"  [tex]y'=75x^4+30x^2+6x[/tex]

Step-by-step explanation:

Part 1.

Given [tex]u=3x^2+2[/tex], find [tex]\frac{du}{dx} \text{ or } u'[/tex].

[tex]u=3x^2+2[/tex]

Apply a derivative to both sides...

[tex]u'=(3x^2+2)'[/tex]

Derivatives of a sum are the sum of derivatives...

[tex]u'=(3x^2)'+(2)'[/tex]

Scalars factor out of derivatives...

[tex]u'=3(x^2)'+(2)'[/tex]

Apply power rule for derivatives (decrease power by 1; mutliply old power as a factor to the coefficient); Derivative of a constant is zero...

[tex]u'=3(2x)+0[/tex]

Simplify...

[tex]u'=6x[/tex]

So, option "a"

Part 2.

Given [tex]v=5x^3+1[/tex], find [tex]\frac{dv}{dx} \text{ or } v'[/tex].

[tex]v=5x^3+1[/tex]

Apply a derivative to both sides...

[tex]v'=(5x^3+1)'[/tex]

Derivatives of a sum are the sum of derivatives...

[tex]v'=(5x^3)'+(1)'[/tex]

Scalars factor out of derivatives...

[tex]v'=5(x^3)'+(1)'[/tex]

Apply power rule for derivatives (decrease power by 1; mutliply old power as a factor to the coefficient); Derivative of a constant is zero...

[tex]v'=5(3x^2)+0[/tex]

Simplify...

[tex]v'=15x^2[/tex]

So, option "d"

Part 3.

Given [tex]y=(3x^2+2)(5x^3+1)[/tex]

[tex]\text{Then if } u=3x^2+2 \text{ and } v=5x^3+1, y=u*v[/tex]

To find [tex]\frac{dy}{dx} \text{ or } y'[/tex], recall the product rule:  [tex]y'=uv'+u'v[/tex]

[tex]y'=uv'+u'v[/tex]

Substituting the expressions found from above...

[tex]y'=(3x^2+2)(15x^2)+(6x)(5x^3+1)[/tex]

Apply the distributive property...

[tex]y'=(45x^4+30x^2)+(30x^4+6x)[/tex]

Use the associative and commutative property of addition to combine like terms, and rewrite in descending order:

[tex]y'=75x^4+30x^2+6x[/tex]

So, option "b"

Use the definitions of expectation and variance.

[tex]E(X) = \displaystyle \int_{-\infty}^\infty x f_X(x) \, dx = \frac14 \int_0^\infty x e^{-x/4} \, dx[/tex]

Integrate by parts,

[tex]\displaystyle \int_a^b u \, dv = uv \bigg|_a^b - \int_a^b v \, du[/tex]

with

[tex]u = x \implies du = dx \\\\ dv = e^{-x/4} \, dx \implies v = -4 e^{-x/4}[/tex]

Then

[tex]E(X) = \displaystyle \frac14 \left(\left(-4x e^{-x/4}\right)\bigg|_0^\infty + 4 \int_0^\infty e^{-x/4} \, dx\right)[/tex]

[tex]E(X) = \displaystyle \int_0^\infty e^{-x/4} \, dx = \boxed{4}[/tex]

(since the integral of the PDF is 1, and this integral is 4 times that)

[tex]V(X) = E\bigg((X - E(X))^2\bigg) = E(X^2) - E(X)^2[/tex]

Compute the so-called second moment.

[tex]E(X^2) = \displaystyle \int_{-\infty}^\infty x^2 f_X(x)\, dx = \frac14 \int_0^\infty x^2 e^{-x/4} \, dx[/tex]

Integrate by parts, with

[tex]u = x^2 \implies du = 2x \, dx \\\\ dv = e^{-x/4} \, dx \implies v = -4 e^{-x/4}[/tex]

Then

[tex]E(X^2) = \displaystyle \frac14 \left(\left(-4x^2 e^{-x/4}\right)\bigg|_0^\infty + 8 \int_0^\infty x e^{-x/4} \, dx\right)[/tex]

[tex]E(X^2) = 8 E(X) = 32[/tex]

and the variance is

[tex]V(X) = 32 - 4^2 = \boxed{16}[/tex]

Answer:

-12

Step-by-step explanation:

Order of operations:

PEMDAS

First, do parenthesis, 2 x 2.2 = 4.4 then do -7.6 - 4.4 = -12