The graph of a system of inequalities shown

The figure second represents the foundation drawing of a provided front, right, and top view option second is correct.

Three-dimensional objects can be represented in two dimensions via an orthographic projection.

As we have given orthographic drawings.

The front, right, and the top view is shown in the picture of the cuboid if merge all the view we will find the three-dimensional figure is a cuboid.

We can represent the cuboid by 2×3 grids as shown in the picture second.

Thus, the figure second represents the foundation drawing of a provided front, right, and top view option second is correct.

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x-intercept -4.8

y-intercept 12

slope 2.5

plot points

x y

-2, 7

-1, 9.5

0, 12

1, 14.5

2 ,17

Answer:

no solutions

Step-by-step explanation:

we use the quadratic formula.

for an equation ax²+bx+c=0, the solutions for x are:

x = (-b±sqrt(b²-4ac))/2a

in our case, a=2, b=-6 and c=5, so:

x= (6±sqrt(36-40))/4

we see that we get a negative number inside the sqrt, so we have no real solutions

Answer:

No solution

Step-by-step explanation:

1) Use the quadratic formula

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

Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.

[tex]2x^2-6x+5=0[/tex]

[tex]a=2\\b=-6\\c=5[/tex]

[tex]x=\frac{-(-6)\pm\sqrt{(-6)^2-4*2*5} }{2*2}[/tex]

2) Simplify

Evaluate the exponent:

[tex]x=\frac{6\pm\sqrt{(-6)^2-4*2*5} }{2*2}[/tex]

[tex]x=\frac{6\pm\sqrt{36-4*2*5} }{2*2}[/tex]

Multiply the numbers:

[tex]x=\frac{6\pm\sqrt{36-4*2*5} }{2*2}[/tex]

[tex]x=\frac{6\pm\sqrt{36-40} }{2*2}[/tex]

Subtract the numbers:

[tex]x=\frac{6\pm\sqrt{36-40} }{2*2}[/tex]

[tex]x=\frac{6\pm\sqrt{-4} }{2*2}[/tex]

Multiply the numbers

[tex]x=\frac{6\pm\sqrt{-4} }{2*2}[/tex]

[tex]x=\frac{6\pm\sqrt{-4} }{4}[/tex]

3) No real solutions because the discriminant is negative

The square root of a negative number is not a real number

[tex]d=-4[/tex]

Result

No solution

The linear equation is y = (1/7)*x + 11/7.

The general linear equation is:

y = a*x + b

Where a is the slope.

Here we know that a = (1/7), then the line is:

y = (1/7)*x + b

We also know that the line contains the point (-4, 1), this means that when x = -4, we must have y = 1.

Replacing that, we get:

1 = (1/7)*(-4) + b

1 + 4/7 = b

7/7 + 4/7 = b

11/7 = b

So the linear equation is:

y = (1/7)*x + 11/7.

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The equation of the linear function of the line is: y = 1/7x + 11/7.

A linear function equation is modelled as, y = mx + b, where m is the slope and b is the y-intercept.

Slope (m) = 1/7, and the line passes through (-4, 1), therefore, substitute (x, y) = (-4, 1) and m = 1/7 into y = mx + b:

1 = 1/7(-4) + b

1 = -4/7 + b

1 + 4/7 = b

11/7 = b

b = 11/7

Plug in the values of m and b into y = mx + b:

y = 1/7x + 11/7

The equation of the linear function is: y = 1/7x + 11/7

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Answer:

okkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

will you be my friend

Step-by-step explanation:

.......... okkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

Answer:  [tex]h(x)=\sqrt{x+4}[/tex]

Step-by-step explanation:

The graph of h(x) is the graph of f(x) shifted 4 units left.

Answer:

The answer is A

Step-by-step explanation:

They both have a maximum of 5!

Answer:

[tex]\fbox {See below} \downarrow[/tex]

Step-by-step explanation:

Question 1 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(4x^{3})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 4(3)x^{3-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 12x^{2}}[/tex]

Question 2 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(\frac{2}{x^{4}})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(2x^{-4})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 2(-4)x^{-4-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = -8x^{-5}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = -\frac{8}{x^{5}}}[/tex]

Question 3 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(3\sqrt{x})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(3(x)^{\frac{1}{2}})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 3(\frac{1}{2})x^{\frac{1}{2}-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{3}{2}x^{-\frac{1}{2}}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{3}{2\sqrt{x}}}[/tex]

Question 4 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(\frac{3}{\sqrt{x}})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(3x^{-\frac{1}{2}})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 3(-\frac{1}{2})x^{-\frac{1}{2}-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = -\frac{3}{2}x^{-\frac{3}{2}}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = -\frac{3}{2x\sqrt{x}}}[/tex]

Question 5 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(\frac{x^{3}}{4})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{1}{4}(3)x^{3-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = \frac{3}{4}x^{2}}[/tex]

Question 6 :

[tex]\mathsf {\frac{dy}{dx} = \frac{d}{dx}(3x^{5})}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 3(5)x^{5-1}}[/tex]

[tex]\mathsf {\frac{dy}{dx} = 15x^{4}}[/tex]

Answer:

Θ = 39°

Step-by-step explanation:

OA = OB ( radii of the circle ) , then

Δ AOB is isosceles with base angles congruent , that is

∠ ABO = ∠ BAO = 52°

the exterior angle of a triangle is equal to the 2 opposite interior angles.

∠ DOB is an exterior angle of the triangle , so

∠ DOB = ∠ ABO + ∠ BAO = 52° + 52° = 114°

the angle between a tangent and the radius at the point of contact is 90°, then

∠ OBC = ∠ ODE = 90°

the sum of the 4 interior angles of quadrilateral OBCD = 360°

∠ OBC + ∠ DOB + ∠ ODC + ∠ BCD = 360°

90° + 114° + (90 + 27)° + Θ = 360°

90° + 114° + 117° + Θ = 360°

321° + Θ = 360° ( subtract 321° from both sides )

Θ = 39°

Step-by-step explanation:

if cospheta =3/5 and pheta is located in the fourth quadrant, then sin2pheta=_____

-8/25

24/5

-24/25

8/25

The correct answer for this question is 8/25

The angle of depression from the top of the lighthouse to the sinking boat is 6.9°.

Trigonometry deals with the relationship between the sides and angles of a right-angle triangle.

From ΔABC it is obtained that;

Height of lighthouse, AC = 25 meters

Distance of boat from the lighthouse, BC = 204 meters

The angle of depression from the top of the lighthouse to the sinking boat is;

[tex]\rm tan \theta = \frac{P}{B} \\\\ \rm tan \theta = \frac{AC}{BC} \\\\ tan \theta = \frac{25}{204} \\\\ \theta = tan^{-1}(0.12)\\\\ \theta = 6.9 ^0[/tex]

Hence, the angle of depression from the top of the lighthouse to the sinking boat is 6.9°

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See attached images for work.

perimeter of octagon

Now area

Area of square

Area of shaded region

Answer:

C.  647 square units

Step-by-step explanation:

To find the shaded area, subtract the area of the unshaded square from the area of the octagon.

Area of the octagon

[tex]\textsf{Area of a regular polygon}=\dfrac{n\:l\:a}{2}[/tex]

where:

Given:

Substitute the given values into the formula and solve for A:

[tex]\implies \textsf{Area}=\sf \dfrac{8 \cdot 13 \cdot 15.69}{2}[/tex]

[tex]\implies \textsf{Area}=\sf \dfrac{1631.76}{2}[/tex]

[tex]\implies \textsf{Area}=\sf 815.88\:\:square \:units[/tex]

Area of the square

[tex]\implies \textsf{Area}=\sf 13^2=169 \:\:square \:units[/tex]

Area of the shaded region

= area of the octagon - area of the square

= 815.88 - 169

= 646.88

= 647 square units (nearest square unit)

The distance between the point (3, 5) and a line with the equation 5x – 3y + 10 = 0 will be 10 / √(34) units.

Let (x₁, y₁) be the point and ax + by + c = 0 be the line.

Then the distance between a point and a line will be

[tex]\rm d = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}[/tex]

The point (3, 5) and a line with the equation 5x – 3y + 10 = 0.

Then the distance between them will be

[tex]\rm d = \dfrac{|5 *3 - 3 *5 + 10|}{\sqrt{5^2 + (-3)^2}}\\d = \dfrac{10 }{\sqrt{34}} \ units[/tex]

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Answer:

h = A / ( 1/2 a - 1/2 b)

Step-by-step explanation:

A = ½ ah - ½ bh

Factor out an h

A = h( 1/2 a - 1/2 b)

Divide each side by ( 1/2 a - 1/2 b)

A / ( 1/2 a - 1/2 b) = h ( 1/2 a - 1/2 b)/ ( 1/2 a - 1/2 b)

A / ( 1/2 a - 1/2 b) = h

h = A / ( 1/2 a - 1/2 b)

Answer:

48 Square Centimeters

Step-by-step explanation:

The red triangular exterior pieces are all equilateral triangles with some side length.

We don't need to know what that side length is. It doesn't matter.

The area of one red equilateral triangle is some area A.

There are 6 of these red triangles, so the red exterior triangular parts combine to a total area of 6*A.

The hexagon is composed of equilateral triangles as well.

Each of these blue equilateral triangles is congruent to any outer red triangle because the side length is the same.

Therefore, the 6 blue equilateral triangles composing this hexagon combine to get a total area of 6*A

The area of the star overall is 12*A because we have 6 blue triangles combining with the 6 red triangles.

There is a total of 12 triangles.

Answer:

452/2 = 226

Step-by-step explanation:

45 divided by 2 is 22 and the remainder would be 12, which could be divided by 2 again to get 6. Add this to the total and you get 226.

Hope this helps! :D

The domain of the given graph is [−3, ∞) and the range is (−∞, 4].

We need to find the domain and range of the given graph.

The range of values that we are permitted to enter into our function is known as the domain of a function. The x values for a function like f make up this set (x). A function's range is the collection of values that it can take.

We can observe that the graph extends horizontally from −3 to the right without a bound, so the domain is [−3, ∞). The vertical extent of the graph is all range values 4 and below, so the range is (−∞, 4].

Therefore, the domain of the given graph is [−3, ∞) and the range is (−∞, 4].

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Answer:

don't know but please keep trying up

Figure it out..............................................

The expected enrolment in 7 years is  1845.

The formula for calculating future value of the number of students is an exponential equation with this form:

FV = P (1 + r) ^n

1500 x (1.03)^7 = 1845

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The rule that describes the composition of transformations that maps figure PQRS to figure P"Q"R"S"is r y-axis • RO, 90° (x, y)

See attachment for complete question

From the image, we have:

P = (1,1)

Q = (1,5)

R = (3,5)

S = (3,1)

Start by reflecting PQRS across the y-axis. i.e. r y-axis

This is done by

(x, y) -> (-x,y)

So, we have:

P' = (-1,1)

Q' = (-1,5)

R' = (-3,5)

S' = (-3,1)

Next, we rotate by 90 degrees counterclockwise i.e. RO, 90° (x, y)

This is done by

(x, y) -> (-y, x)

So, we have:

P'' = (-1,-1)

Q'' = (-5,-1)

R'' = (-5,-3)

S'' = (-1,-3)

Hence, the rule that describes the composition of transformations that maps figure PQRS to figure P"Q"R"S"is r y-axis • RO, 90° (x, y)

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Answer and Step-by-step explanation:

There is a cube with side length 3. The volume of this cube is 27.

Answer:

7+45/n or 5

Step-by-step explanation:

45 divided by n or technically 5 is 9 plus 7 is 16 :)

Answer:7+45/n or 5

Step-by-step explanation:

Answer:  2/6+5/6 + 1.167

Answer:

1.16

Step-by-step explanation:

Step-by-step explanation:

the area of a triangle is

baseline × height / 2

baseline and height must be at a right angle with each other.

so, in our case

6 × 11 / 2 = 3 × 11 = 33 cm²

Answer:

Area = 33 cm²

Step-by-step explanation:

[tex]\textsf{Area of a triangle}=\sf \dfrac{1}{2}bh[/tex]

where:

If we turn the triangle counterclockwise, so that the side that measures 6 cm is the base, we will see that the length that measures 11 cm is the height of the triangle (see attached).

Therefore:

Substitute the found values into the formula and solve for A:

[tex]\implies \sf A=\dfrac{1}{2}(6)(11)[/tex]

[tex]\implies \sf A=\dfrac{1}{2}(66)[/tex]

[tex]\implies \sf A=33[/tex]

Therefore, the area of the triangle is 33 cm²

The z-score of an appliance that stopped working at 64 months is 2.

The mean of observations is equal to the ratio of sum of all the observations and the number of observations.

Given, the set of life spans of an appliance is normally distributed with a mean mu = 48 months and a standard deviation sigma = 8 months

The z-score is then calculated as

z = x - μ /σ

For an appliance that stopped working at 64 months, we have

x = 64

On substituting the values, we get

z = 64-48/8

z = 2

Hence, the z-score of an appliance that stopped working at 64 months is 2

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Answer:

D on edge

Step-by-step explanation:

2