Oh for the love of noodles, please help me T-T

Zolol [24]

Surface area = sum of 2 triangles + sum of the 3 rectangles

= 2 * 84 + 18*24 + 18*25 + 18*7 = 1176 ft^2

volume = 1/2*7 *24 * 18 = 1512 ft^3
surface area to volume = 1176:1512 = 0.78 which is less than 1.
so the company wil manufacture it.

5 0

3 years ago

The accompanying diagram show a cross-section of a rectangular pyramid. The cross sectional area is 36in and is 3 inches from th

LiRa [457]

The area of the rectangular base is the amount of space on the rectangular base

The area of the rectangular base is 60 square inches

<h3>How to determine the area of the rectangular base?</h3>

The question is incomplete; as the diagram is not given.

So, I will apply the concept of similar shapes to determine the area of the rectangular base

To determine the area of the rectangular base, we make use of the following equivalent ratio:

Ratio = Height : Area

This gives

3 : 36 = 5 : Area

Express as fraction

36/3 = Area/5

Evaluate the quotient

12 = Area/5

Multiply both sides by 5

Area = 60

Hence, the area of the rectangular base is 60 square inches

Read more about areas at:

brainly.com/question/24487155

3 0

2 years ago

The equation 3x + 2 = 29 is modeled below.

DanielleElmas [232]

Answer:

Subtract 2 from both sides.

Step-by-step explanation:

Remember to do the opposite of PEMDAS. PEMDAS is the order of operation, and stands for:

Parenthesis

Exponents (& Roots)

Multiplication

Division

Addition

Subtraction

> Also, note the equal sign, what you do to one side, you do to the other side of the equation.

First step is to subtract 2 from both sides of the equation:

3x + 2 (-2) = 29 (-2)

3x = 29 - 2

3x = 27

The second step is to divide 3 from both sides of the equation:

(3x)/3 = (27)/3

x = 27/3

x = 9

~

7 0

2 years ago

Read 2 more answers

The equation of a circle is (x + 6)^2 + (y - 4)^2 = 16. The point (-6, 8) is on the circle.

pantera1 [17]

Answer:

y = 8 is the equation of tangent.

Step-by-step explanation:

The equation of the tangent to the circle at (-6,8) is of the form:

y = mx + c

where m is the slope of the tangent and c is the y-intercept.

The point (-6,8) lies on the circle and the tangent line as well.

Hence (-6,8) satisfies the line equation:

8 = m(-6) + c ⇒ c-6m = 8 -------------1

We know that slope of two perpendicular lines are related as:

$m_{1}\times m_{2}=-1$

At any point on the circle, the normal line at a point is always perpendicular to the tangent line at that point.

Hence :

$m_{normal} \times m_{tangent}=-1$

We can find the slope of the normal at point (-6,8) as it passes through the centre of the circle (-6,4) by using the two-points formula for slope.

$m=\frac{y_2-y_1}{x_2-x_1}$

$=\frac{8-4}{-6+6}$

= ∞

Slope of the normal is infinity and hence slope of tangent is -1/∞ = 0

Hence m=0

Putting m=0 in equation 1 we get:

c = 8

The equation of tangent line at (-6,8) is:

y = 8

6 0

3 years ago

PLEASE OPEN THIS, I NEED HELP!!!! WHAT BELONGS IN THE ? AND WHY?

ruslelena [56]

Answer: see below

<u>Step-by-step explanation:</u>

Write each equation in y = mx + b format where m is the slope and b is the y-intercept.

Left side: -4 ≤ x < -1

If you continue the line through the y-axis it will pass through (0, 4) --> b = 4

The rise over run is -1 over 1 --> m = -1

y = (-1)x + (4) --> y = -x + 4

Right side: -1 < x < 4

The line passes through (0,0) --> b = 0.

The rise over run is -1 over 1 --> m = -1

y = (-1)x + (0) --> y = -x

$\large\boxed{f(x)=\bigg\{\begin{array}{ll} -x+4&;-4\leq x$

5 0

3 years ago