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3 years ago

What is 5 (p-10) expanded

Mathematics

2 answers:

Andru [333]3 years ago

7 0

Answer:

(5 + P) - (5 + 10)

Step-by-step explanation

i think sorry if i'm wrong

V125BC [204]3 years ago

3 0

Answer:

5p - 50

You distribute

5 x p
5 x -10

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PLSSS HELP GIVING 5 POINTS PLUS BRAINLIEST

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Answer:The answer should be D based on the data given.

Step-by-step explanation:

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3 years ago

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What is the value of x

Shkiper50 [21]

Answer:

124°.

Step-by-step explanation:

x+24°+32°=180° (sum of angles is 180°)

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3 years ago

What is (2/x^2)^3 reduced to simplest form ? A. 8x^-6 B. 8x^6 C. 6/x^6 D. 8/ x^6

pashok25 [27]

<h2>What is (2/x^2)^3 reduced to simplest form ?</h2>

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2 years ago

prove that if f is integrable on [a,b] and c is an element of [a,b], then changing the value of f at c does not change the fact

Neko [114]

Answer with Step-by-step explanation:

We are given that if f is integrable on [a,b].

c is an element which lie in the interval [a,b]

We have to prove that when we change the value of f at c then the value of f does not change on interval [a,b].

We know that limit property of an integral

$\int_{a}^{b}f dt=\int_{a}^{c}fdt+\int_{c}^{b} fdt$

$\int_{a}^{b} fdt=f(b)-f(a)$ ....(Equation I)

Using above property of integral then we get

$\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b} fdt$ ......(Equation II)

Substitute equation I and equation II are equal

Then we get

$\int_{a}^{b}fdt= f(c)-f(a)+{f(b)-f(c)}$

$\int_{a}^{b}fdt=f(c)-f(a)+f(b)-f(c)=f(b)-f(a)$

$\int_{a}^{c}fdt+\int_{c}^{b}fdt=f(b)-f(a)$

Therefore, $\int_{a}^{b}fdt=\int_{a}^{c}fdt+\int_{c}^{b}fdt$ .

Hence, the value of function does not change after changing the value of function at c.

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3 years ago

Write an equation to model growth or decay in the following scenario. A stock is declining at a rate of 25% of its value every 2

yawa3891 [41]

The exponential function that models this situation is:

$y(t) = 225(0.75)^{\frac{t}{2}}$

------------------------------------------

A decaying exponential function has the following format:

$y(t) = y(0)(1 - r)^t$

In which:

y(0) is the initial value.
r is the decay rate, as a decimal.

In this problem:

The stock started at $225, thus $y(0) = 225$ .
Declines at a rate of 25% every 2 weeks, thus $r = 0.25, t = \frac{t}{2}$

The equation is:

$y(t) = y(0)(1 - r)^t$

$y(t) = 225(1 - 0.25)^{\frac{t}{2}}$

$y(t) = 225(0.75)^{\frac{t}{2}}$

A similar problem is given at brainly.com/question/24282972

3 0

3 years ago

What is 5 (p-10) expanded​

What is 5 (p-10) expanded