Question 2 of 11
What system of equations would you use to solve the problem below?
A test is worth 100 points. Multiple-choice questions are worth 3 points and
short-answer questions are worth 5 points. If the test has 28 questions, how
many multiple-choice questions are there?

Answer:

expression for (g + h)([tex]x\\[/tex]) = [tex]2x[/tex] + [tex]3x^2\\[/tex] and for (g - h)([tex]x[/tex]) = [tex]2x[/tex] - [tex]3x^2[/tex]

value of (g .h)(2) = 24

Step-by-step explanation:

function operations: Functions with overlapping domains can be added, subtracted, multiplied and divided. If f(x) and g(x) are two functions, then for all x in the domain of both functions we can find the sum, difference, product and quotient

given :

g([tex]x[/tex]) = [tex]2x[/tex]

h([tex]x[/tex]) = [tex]3x^2[/tex]

we know that ;

(g + h)([tex]x[/tex]) = g([tex]x[/tex]) + h([tex]x[/tex])

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

similarly;

(g - h)([tex]x[/tex]) = g([tex]x[/tex]) - h([tex]x[/tex])

             =  [tex]2x[/tex] - [tex]3x^2[/tex]

now,  

(g . h)([tex]x[/tex]) = 2([tex]3x^2\\[/tex])

             = 6[tex]x^2\\[/tex]

for (g .h)(2) = 6([tex]2^2[/tex])

                  = 6 × 4

                  = 24

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The cosine ratios which are correct for the ΔPQR are CosineP = r/q and CosineR = p/q.

We have,

Δ PQR,

Length of hypotenuse PR = q

The length of PQ = r,

And the length of QR = p

NOw,

We know that Trigonometric ratios of CosineΘ;

CosineΘ = Base/Hypotenuse

So,

CosineP = r/q

And,

CosineR = p/q

So, from the above find out values we can say that these values are given in option (a) and option (d).

Therefore, the cosine ratios which are correct for the ΔPQR are CosP = r/q and CosR = p/q.

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Answer: 54%

Step-by-step explanation:

You will need to divide these numbers.

[tex]6/11=0.54[/tex]

After you divide that, simplify that to a percentage by multiplying it by 100.

[tex]0.54 *100=54%[/tex]%

I hope this helps!!

Answer:

1 1/2

Step-by-step explanation:

Convert both fractions into decimals

⇒ 3/4 (0.75) and 1/2 (0.50)

Divide the decimals:

⇒0.75 ÷ 0.50

Find the quotient

⇒ 1.5

Convert the quotient to a fraction

⇒ 1 1/2

Terms:

Operations with fractions

Answer:

C. 4x²√(3x)

Step-by-step explanation:

if x > 0

[tex]\sqrt{48x^{5}} =\sqrt{48} \times \sqrt{x^{5}}[/tex]

          [tex]=\sqrt{16\times 3} \times \sqrt{x^{4}\times x}[/tex]

          [tex]=\sqrt{16} \times \sqrt{3} \times \sqrt{x^{4}} \times \sqrt{x}[/tex]

          [tex]=4\times \sqrt{3} \times \sqrt{\left( x^{2}\right)^{2} } \times \sqrt{x}[/tex]

          [tex]=4\times \sqrt{3} \times x^{2}\times \sqrt{x}[/tex]

          [tex]=4\times x^{2} \times \sqrt{3} \times \sqrt{x}[/tex]

          [tex]=4\times x^{2}\times \sqrt{3x}[/tex]

Answer:

d. 11:15

Step-by-step explanation:

1) officer1: distance 75 miles, velocity 60mil/h; officer2: distance 25 miles, velocity 55mil/h;

2) time, which is spent officer1 is: 75/60=1.25[h]=75 [min]. If the officer1 leaves his office at 10:00, the arrival time is 11:15;

3) time which is spent officer2 is: 25/55=5/11[h]≈27[min]. The arrival time is 10:27.

4) finally, the both officers are in the required location at 11:15.

Answer:

y = 1, 9.

Step-by-step explanation:

They are the y values of the points of intersection of the 2 parabolas.

y = 1, 9.

Step-by-step explanation:

x^3 is a perfect cube, 8 is a perfect cube, so we use difference of cubes.

[tex] {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )[/tex]

Cube root of x^3 is x.

Cube root of 8 is 2

So

a=x

b= 2.

[tex](x - 2)( {x}^{2} - 2x + 4)[/tex]

Set these equations equal to zero

[tex]x - 2 = 0[/tex]

[tex]x = 2[/tex]

[tex] {x}^{2} - 2x + 4 = 0[/tex]

If we do the discriminant, we get a negative answer so we would have two imaginary solutions,

Thus the only real root is 2.

If you want imaginary solutions, apply the quadratic formula.

[tex]1 + i \sqrt{ 3 } [/tex]

and

[tex]1 - i \sqrt{3} [/tex]

Answer:

[tex] {5}^{ \frac{1}{12} } = \sqrt[12]{5} [/tex]

Step-by-step explanation:

[tex] \frac{ \sqrt[3]{5} }{ \sqrt[4]{5} } = {5}^{ \frac{1}{3} } \div {5}^{ \frac{1}{4} } = {5}^{ \frac{4}{12} } \div {5}^{ \frac{3}{12} } = {5}^{ \frac{1}{12} } = \sqrt[12]{5} [/tex]

THIS IS THE ANSWER: 5 to the power of one-twelfth (Take a look at the image if this does not make sense) :))

:))

:)))   HAPPY SLAPPY!!!!

:))))

"UPDATE" : This was correct on my test. Also I realized to see the image in this answer keep your cursor on the side of your screen, not on the image :D

The equation for the red graph is f(x) = -x².

From the information given, g(x) = 4 - x².

The vertex of the point (0, 4) and (0, 0) for g(x) and f(x) respectively. The rule of the translation will be:

(x, y) = (x, y -4).

The equation of the function will be:

f(x) = g(x) - 4

f(x) = (4 - x²) - 4

f(x) = -x²

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Answer:

B

Step-by-step explanation:

when x is 0 y=4(0)+12 y=12

y is 0 y=4(-3) +12 = 12=12

Answer:

A.

Step-by-step explanation:

The parabola opens down, so the "a" value (the number in front of x^2) must be negative. The y-intercept is (0, 5) so the constant must be positive 5.

It is not okay to conduct the simulation because people generally favor some numbers.

From the information given, the student wants to simulate 37 birthdays, but she does not have a calculator or software program available.

In this case, it's not okay to conduct the simulation because people generally favor some numbers.

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Answer:

[tex]\dfrac{1}{3}[/tex]

Step-by-step explanation:

Given:

[tex]\displaystyle \sum^{\infty}_{n=1} 4^{-n}[/tex]

The sigma notation means to find the sum of the given geometric series where the first term is when n = 1 and the last term is when n = ∞.

To find the first term in the series, substitute n = 1 into the expression:

[tex]\implies a_1=4^{-1}=\dfrac{1}{4}[/tex]

The common ratio of the geometric series can be found by dividing one term by the previous term:

[tex]\implies r=\dfrac{a_2}{a_1}=\dfrac{4^{-2}}{4^{-1}}=\dfrac{1}{4}[/tex]

As | r | <  1  the series is convergent.

When a series is convergent, we can find its sum to infinity (the limit of the series).

Sum to infinity formula:

[tex]S_{\infty}=\dfrac{a}{1-r}[/tex]

where:

Substitute the found values of a and r into the formula:

[tex]S_{\infty}=\dfrac{\frac{1}{4}}{1-\frac{1}{4}}=\dfrac{1}{3}[/tex]

Therefore:

[tex]\displaystyle \sum^{\infty}_{n=1} 4^{-n}=\dfrac{1}{3}[/tex]

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Answer:

x² + 4x + 3

---------------

       2

---- => fraction line

x² + 12 is equivalent to the polynomial (x + 2[tex]\sqrt{3}[/tex]i) (x + 2[tex]\sqrt{3}[/tex]i).

Polynomials are defined as algebraic expressions consisting of constants, variables and exponents combined using operations like addition, multiplication, subtraction and division.

The given expression is x² + 12.

We have the identity (a + b) (a - b) = a² - b²

(x + 2[tex]\sqrt{3}[/tex]i) (x + 2[tex]\sqrt{3}[/tex]i) = x² - (2[tex]\sqrt{3}[/tex]i)²

                                = x² - (2² × √3² × i²)

We know that i² = -1, where i is the imaginary number in complex numbers

(x + 2[tex]\sqrt{3}[/tex]i) (x + 2[tex]\sqrt{3}[/tex]i) = x² - (4 × 3 × -1)

                                = x² + 12

Hence we get that (x + 2[tex]\sqrt{3}[/tex]i) (x + 2[tex]\sqrt{3}[/tex]i) is equivalent to x² + 12.

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Answer:

B

Step-by-step explanation:

(f + g)(x)

= f(x) + g(x)

= x + 8 - 4x - 3 ← collect like terms

= - 3x + 5

Answer:

[tex]\huge\boxed{\sf (f + g)(x) = -3x + 5}[/tex]

Step-by-step explanation:

Add both functions

(f + g)(x) = x + 8 + (-4x - 3)

(f + g)(x) = x + 8 - 4x - 3

(f + g)(x) = x - 4x + 8 - 3

(f + g)(x) = -3x + 5

[tex]\rule[225]{225}{2}[/tex]

Answer:

$12.

Step-by-step explanation:

1) if price of one doll is 'x', of one dress is 'y', then the condition 'buys one doll and one dress, it will cost her $21.00' can be written as x+y=21;

2) the condition 'buys two dolls and three dresses, it will cost her $51.00' can be written as 2x+3y=51;

3) it is possible to make up the system:

[tex]\left \{ {{x+y=21} \atop {2x+3y=51}} \right. \ = > \ \left \{ {{x=12} \atop {y=9}} \right.[/tex]

4) the price of one doll is 12$.

Answer:

Even though only 6% of students scored higher than Karl, we cannot know his personal score based on his percentile. He could have gotten a 62, as long as everyone else [94% of the class] got an even lower score.

So, we cannot know Karl's mark, we are only able to understand Karl's score in relation to the rest of the students in his law course.

Answer:

Step-by-step explanation:

the numerator goes in side of the radical

Answer:

BC is 22.5 mi

Step-by-step explanation:

Using the pythagorean theorem formula. c² = a² + b²

BC² = AB² + AC²

BC² = 19² + 12²

BC² = 361 + 144

BC² = 505

√BC² = √505

BC = 22.5

Therefore BC is 22.5 mi

(i) The Maclaurin's series is:

1 ₊ x ₊ x²/2! . 1 ₋ 3x⁴/4! ₊ ...........

(ii) Integral of limit o to 1∫ e^sinx is 0

Maclaurin's expansion is as follows:

y(x) = y₀ ₊ x y₁(0) ₊ x²/2! y₂(0) ₊ x³/3! y₃(0) ₊ ........

y = e^sinx

y(0) = e^sin0

y(0) = 1

Differentiate y with respect to x

y₁(x) = e^sinx . cos x

y₁(0) = 1 [i.e y₁ = cosx.y]

y₁(0) = 1

Differentiate y₁ with respect to x

y₂(x) = cosc . y₁ ₋ sinx . y

∴ y₂(0) = 1 ₋ 0 = 1

Differentiate y₂ with respect to x

y₃ = cosx . y₂ ₋ (2 sinx ₊ cosx)y₁

y₃(0) = 1 ₋(0 ₊ 1)1

y₃(0) = 0

Differentiate y₃ with respect to x

y₄ = cosx . y₃ sinx . y₃ ₋ (2sinx ₊ cosx)y₂ ₋ (2cos x ₋ sinx) y₁

y₄(0) = 0 ₋ (3 . 0 ₊ 1)1 ₋ (2 ₋ o) 1

y₄(0) = -3

∴ From maclaurin's equation we get:

e^sinx = 1 ₊ x . 1 ₊ x²/2! . 1 ₊ x³/3! (0) ₊ x⁴/4! (₋3) ₊ ....

           =1 ₊ x ₊ x²/2! . 1 ₋ 3x⁴/4! ₊ ...........

Hence we get the the series expansion as 1 ₊ x ₊ x²/2! . 1 ₋ 3x⁴/4! ₊ ...........

(ii) limit 0 to 1∫e^sinx

integration formula ∫e^x dx = e^x + c

                    limit 0 to 1 ∫e^sinx = [e^sinx + c ]limit 0 to 1

                                                 = e^sin0 ₋ e^sin1

                                                 = 1 ₋ 1

                                                 = 0

Hence we get the answer as 0

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The number best represents the slope of the graphed line is 1/2.

We have given the two points,

(-6,-7) and (4,-2) and slope-intercept is 8 units

[tex]y-y_{1}=m\left(x-x_{1}\right)[/tex]

y = y coordinate of the second point

[tex]y_1[/tex] = y coordinate of point one

m = slope

x = x coordinate of the second point

[tex]x_{1}[/tex]= x coordinate of point one

Therefore the

[tex]slope=\frac{y_2-y_1}{x_2-x_!}[/tex]

[tex]=\frac{-2-(-7)}{4-(-6)} \\=\frac{5}{10}\\ =\frac{1}{2}[/tex]

Therefore the correct option is 1/2.

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Answer:

5/17

Step-by-step explanation:

If we do the multiplication first, we get:

[tex]\frac{4-(12\div-12)}{13+(-8\div-2)}[/tex]

If we do the division next, we get:

[tex]\frac{4+1}{13+4} = 5/17[/tex]

Answer:

f'(1)=-1

Step-by-step explanation:

You can just find the derivative using the power rule for d/dx (x^-1)=-1x^-2=-1/x^2 then find f'(1)=-1.

I used the limit definition of the derivative to find the slope in the picture attached

Answer:

u should add 118/9

Step-by-step explanation:

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According to the FIFO method, the ending inventory balance for Commodity Z is $1,405

FIFO means that earlier inventory is sold first and later inventory would be the last goods left.

Since there were 23 units left, all 16 units from the last purchase and 7 units from the second purchase would be available:

= (16 x 62) + (7 x 59)

= 992 + 413

= $1,405

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