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2 years ago

PEEPS PLEASE HELP PLEASE AND THANK YOU

Mathematics

1 answer:

Feliz [49]2 years ago

4 0

Answer:

The correct answer would be C) {28}

Step-by-step explanation:

Step 1, Cross multiply

7/p x 8/2

= 56= p x 2

Step 2: Switch the sides

p x 2 = 56

Step 3: Divide both sides by 2

( remember that we cross multiplied.)

p x 2/2 = 56/2

Step 4: Simplify

p = 28

So that answer would be C. p = 28

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Find the differential of each function. (a) y = tan( 5t ) dy = Correct: Your answer is correct. (b) y = 5 − v2 5 + v2

Trava [24]

Answer:

$(a)$ $dy = 5sec^2(5t) \ dt$

$(a) \ dy = \frac{-20v}{(5+v^2)^2} \ dt$

Step-by-step explanation:

Given;

(a) y = tan(5t)

$(b) \ y = \frac{5-v^2}{5+v^2}$

Solving for (a)

y = tan(5t)

let u = 5t

⇒y = tan(u)

du/dt = 5

dy/du = sec²u

$\frac{dy}{dt} =\frac{dy}{du} *\frac{du}{dt} \\\\\frac{dy}{dt} = sec^2(u)*5\\\\\frac{dy}{dt} = 5sec^2(u)\\\\\frac{dy}{dt} = 5sec^2(5t)$

$dy = 5sec^2(5t) \ dt$

Solving for b;

let u = 5 - v²

du/dv = -2v

let v = 5+ v²

dv/du = 2v

$\frac{dy}{dv} = \frac{vdu - udv}{v^2} \\\\\frac{dy}{dv} = \frac{-2v(5+v^2) - 2v(5-v^2)}{(5+v^2)^2}\\\\\frac{dy}{dv} = \frac{-10v-2v^3-10v+2v^3}{(5+v^2)^2}\\\\$

$\frac{dy}{dv} = \frac{-10v-10v}{(5+v^2)^2}\\\\\frac{dy}{dv} = \frac{-20v}{(5+v^2)^2}\\\\dy = \frac{-20v}{(5+v^2)^2} dt$

4 0

3 years ago

Solving Two-Step Inequalities

Nuetrik [128]

Answer:

x > $\frac{9}{5}$

Step-by-step explanation:

Given

$\frac{2}{3}$ x - $\frac{1}{5}$ > 1

Multiply through by 15 to clear the fractions

10x - 3 > 15 ( add 3 to both sides )

10x > 18 ( divide both sides by 10 )

x > $\frac{18}{10}$ , that is

x > $\frac{9}{5}$

6 0

4 years ago

Evaluate 13+ 6/y when y=6

liubo4ka [24]

Answer: 14

Step-by-step explanation:

13 + 6/6 = ?

13 + 1 = 14

3 0

3 years ago

A farmer owns 24 acres of land. He plans to use 6 acres for an entrance into the farm and partition the remaining land into 1/3

alex41 [277]

Answer:

he will have 54 1/3 acres of land

Step-by-step explanation:

8 0

3 years ago

each side of a small square gym is 17 meters long. a square classroom has half the area of the gym. how long is each side of the

Semmy [17]

If the gym is a square with 17 meters sides, its area is given (as any other square) by the square of the side:

$A_{gym} = 17^2 = 289 \text{ m}^2$

So, the area of the classroom is

$A_{class} = \dfrac{289}{2}=144.5 \text{ m}^2$

We can deduce the side of the classroom only assuming that it is a square as well: if this is the case, the side of a square with given area is the square root of the area, so we have

$s_{class} = \sqrt{144.5} \approx 12.02 \text{ m}$

5 0

3 years ago