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4 years ago

9

Graham joined two congruent square pyramids to form the composite solid. If the lateral faces of the pyramids each have an area

of 18.4 cm2, what is the total surface area of the composite solid?

2 answers:

erma4kov [3.2K]4 years ago

6 0

Your answer is B because if you add 18.4 times the lateral faces you get 110.4

Brrunno [24]4 years ago

4 0

Answer:

B ) 110.4

Step-by-step explanation:

You might be interested in

A car wheel has a radius of 16 inches. Through what angle ( to the nearest tenth of a degree ) does the wheel turn when the car

olya-2409 [2.1K]

First of all, let's convert all the measures to the same unit: 4 feet are 48 inches.

Now, as the wheel turns, there is a proportion between the angle and the distance travelled: for example, when the car moves forward a whole circumference, the angle will be 360°. Conversely, if the wheel turns 180°, then the car will move forward a distance which is half the circumference of the wheel, and so on.

Since the radius is 16 inches, the circumference will be

$C=2\pi r = 32\pi$

So, we have the following proportion:

$360\div 32\pi = x \div 48$

that you can read as: "if an angle of 360 corresponds to a distance travelled of $32\pi$ , then the unknown angle x corresponds to a distance travelled of 48 inches.

Solving for x, we have

$x = \dfrac{360\cdot 48}{32\pi} = \dfrac{17280}{32\pi} = 171.887338539\ldots \approx 171.9$

5 0

4 years ago

The cube of the difference of 4 times x and 9

grigory [225]

Answer:

<h2>64x³ - 432x² + 972x - 729</h2>

Step-by-step explanation:

Write the statement into an expression

<h2>(4x - 9)³</h2><h2 /><h3>Now multiply that out...</h3><h3 /><h3>(4x - 9)²(4x - 9)</h3><h3 /><h3> Foil the first binomial</h3><h3 /><h3> (16x² - 72x + 81)(4x - 9)</h3><h3 /><h3>Multiply the two polynomials together</h3><h3 /><h3> 16x²(4x) - 72x(4x) + 81(4x) + 16x²(-9) - 72x(-9) + 81(-9)</h3><h3 /><h3> 64x³ - 288x² + 324x - 144x² + 648x - 729</h3><h3 /><h3>combine like terms...</h3><h3 /><h3>64x³ - 432x² + 972x - 729</h3>

8 0

3 years ago

MATH QUESTION: Julie bought a bag of parsnips that weighed 4 2/7 pounds. She also bought a bag of turnips that weighed 1 1/3 tim

kati45 [8]

Answer:

Julie buy $5\frac{5}{7}$ pounds of turnips.

Step-by-step explanation:

We are given that Julie bought a bag of parsnips that weighed ( $4\frac{2}{7}$ ) pounds. She also bought a bag of turnips that weighed ( $1\frac{1}{3}$ ) times as much as the parsnips.

We have to find how many pounds of turnips did Julie buy.

Firstly, the weight of a bag of parsnips that Julie bought = $4\frac{2}{7}$ = $\frac{30}{7}$ pounds

Now, it is stated that she bought a bag of turnips that weighed ( $1\frac{1}{3}$ ) times as much as the parsnips, that means;

Weight turnips bag = $\frac{4}{3} \times \frac{30}{7}$

= $4 \times \frac{10}{7}$

= $\frac{40}{7}$

So, Julie buy $5\frac{5}{7}$ pounds of turnips.

5 0

3 years ago

Match the two numbers with their least common multiple (LCM).

vovangra [49]

Answer:

The middle one cause 8 cant go into 30 and 4 cant go into 30

Step-by-step explanation:

4 0

3 years ago

The expression (secx + tanx)2 is the same as _____.

trapecia [35]

<u>Answer:</u>

The expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

<u>Solution:</u>

From question, given that $\bold{(\sec x+\tan x)^{2}}$

By using the trigonometric identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$ the above equation becomes,

$(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x$

We know that $\sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}$

$(\sec x+\tan x)^{2}=\frac{1}{\cos ^{2} x}+2 \frac{1}{\cos x} \frac{\sin x}{\cos x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

$=\frac{1}{\cos ^{2} x}+\frac{2 \sin x}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

On simplication we get

$=\frac{1+2 \sin x+\sin ^{2} x}{\cos ^{2} x}$

By using the trigonometric identity $\cos ^{2} x=1-\sin ^{2} x$ ,the above equation becomes

$=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}$

By using the trigonometric identity $(a+b)^{2}=a^{2}+2ab+b^{2}$

we get $1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}$

$=\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}$

$=\frac{(1+\sin x)(1+\sin x)}{1-\sin ^{2} x}$

By using the trigonometric identity $a^{2}-b^{2}=(a+b)(a-b)$ we get $1-\sin ^{2} x=(1+\sin x)(1-\sin x)$

$=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}$

$= \frac{1+\sin x}{1-\sin x}$

Hence the expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

8 0

3 years ago