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2 years ago

Find the total surface area of this cuboid. 5 cm 2.3 cm 7.5 cm

Mathematics

1 answer:

Mekhanik [1.2K]2 years ago

8 0

Answer:

The total surface area of this cuboid is 132.5 sq.cm

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In a one-month period, a convenience store

Vera_Pavlovna [14]

So, youd start by adding the two sales together to get total sales. Then you would divide that number by the gas sales to get the percent

65560+36740 = x

36740/ x = the percentage :)

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3 years ago

156 is 60% of what number

Tanzania [10]

The answer is 93.6
156x .6 = 93.6

8 0

3 years ago

Figure a is dilated by a scale factor of 1/2 to form figure b. What is the length of x?

PIT_PIT [208]

Answer:

The correct option is 3. The value of x is 8 inches.

Step-by-step explanation:

It is given that Figure A is dilated by a scale factor of 1/2 to form Figure B. It means both figure are similar and corresponding sides of similar figures are proportional.

$\text{scale factor}=\frac{\text{Side of image}}{\text{Side of preimage}}$

$\frac{1}{2}=\frac{x}{16}$

Multiply both sides by 16.

$\frac{16}{2}=x$

$8=x$

Since scale factor is 1/2, so we can directly say that the side length of image is half of the corresponding side of preimage.

The length of x is 8 inches. Therefore the correct option is 3.

3 0

4 years ago

Read 2 more answers

Steven wishes to save for his retirement by depositing $2,000 at the beginning of each year for thirty years. Exactly one year a

Ad libitum [116K]

Step-by-step explanation:

i = interest 3% for 30 years

This is a simple dynamical system for whom the the solutions are given as

$S=R[\frac{(i+1)^n-1}{i}](i+1)$

putting values we get

S=2000[\frac{(1.03)^{30}-1}{0.03}](1.03)

= $98005.35

withdrawal of money takes place from one year after last payment

To determine the result we use the present value formula of an annuity date

$P = R\frac{1-(1+i)^{-n}}{i}{i+1}$

we need to calculate R so putting the values and solving for R we get

R= $6542.2356

8 0

3 years ago

Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$

5 0

3 years ago