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2 years ago

Of the 400 words that brianna wrote for her book report

Mathematics

1 answer:

Black_prince [1.1K]2 years ago

8 0

The question here is a little unclear. Would love to help but please clarify :)

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spayn [35]

Answer:

$(8 - 3)^2 = 5^2$

Step-by-step explanation:

Given

$(8 - 3)$

Required

Determine the second power

The second power of an expression is the square of the expression.

So, the second power of $(8 - 3)$ is

$(8 - 3)^2$

Solving further:

$(8 - 3)^2 = 5^2$

Hence:

Option (c) is correct

8 0

2 years ago

Find the rate of markup if a book cost the retailer 17.50 and sells for $24.95. Round the the nearest tenth if needed

CaHeK987 [17]

Mark up = profit ÷ cost

Let M = mark up

Let p = profit

Let c = cost

M = p/c

Take it from here.

4 0

3 years ago

Does 6n-18=4(n+2) equal 1n?

Mariulka [41]

Answer:

No.

Step-by-step explanation:

If we were to fill in the slots, the equation would be...

6x3-18=4

6 times three is 18, and 18 - 18 equals 0. So the answer would not be 1.

4 0

2 years ago

Milton makes trail mix for his hiking group he makes a 1 and 1/4 pound of peanuts 14 oz of raisins 12 oz of walnuts in 10 oz of

strojnjashka [21]

Answer:

Each hiker receives 7 ounces of trail mix

Step-by-step explanation:

Trail mix=quantity of peanuts+quantity of raisins+quantity of walnuts+quantity of chocolate chips

where;

Quantity of peanuts=1.25 pounds

Quantity of raisins=14 oz, since 1 pound=16 oz.

Quantity of raisins=(14/16)=0.875 pounds

Quantity of walnuts=12 oz=(12/16)=0.75 pounds

Quantity of chocolate chips=10 oz=(10/16)=0.625 pounds

Replacing;

Trail mix=(1.25+0.875+0.75+0.625)=3.5 pounds

Trail mix=Quantity per hiker×Number of hikers

where;

Trail mix=3.5 pounds

Quantity per hiker=q

Number of hikers=8

Replacing;

3.5=q×8

q=3.5/8=0.4375 pounds

I pound=16 ounces

q=0.4375×16=7 ounces

Each hiker receives 7 ounces of trail mix

8 0

2 years ago

Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):

$y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1 \right)} \, dx$

$y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx$

$y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C$

$C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0$

$C = 0$

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

5 0

3 years ago