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3 years ago

Choose the product.

Mathematics

2 answers:

Grace [21]3 years ago

6 0

Answer is A, -20p^5-15p^4+5p^3

ValentinkaMS [17]3 years ago

3 0

Simply distribute, multiplying the coefficients and adding exponents when necessary...

<span>-5p^3 (4p^2 + 3p - 1)

The first term I'll explain then just do the rest

-20p^5

because -5*4 = -20 and p^3 * p^2 = p^5. Since we are multiplying bases with exponents, we keep the bases and add the exponents, now for the rest of the expression

</span>-20p^5 -15p^4 + 5p^3, Option A.

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Translate the sentence into an equation.

lozanna [386]

9w - 7 = 5

9w= 5+7

9w=12

w= 4/3 or 1.33

3 0

3 years ago

2)<br> Solve the quadratic equation by taking square roots.<br> 4x^2 = 24

Veronika [31]

Answer:

4x²=24

or,x²=24÷4

or,x=root 6

8 0

3 years ago

6. Two observers, 7220 feet apart, observe a balloonist flying overhead between them. Their measures of the

MaRussiya [10]

Answer:

The ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

Step-by-step explanation:

Let's call:

h the height of the ballonist above the ground,

a the distance between the two observers,

$a_1$ the horizontal distance between the first observer and the ballonist

$a_2$ the horizontal distance between the second observer and the ballonist

$\alpha _1$ and $\alpha _2$ the angles of elevation meassured by each observer

S the area of the triangle formed with the observers and the ballonist

So, the area of a triangle is the length of its base times its height.

$S=a*h$ (equation 1)

but we can divide the triangle in two right triangles using the height line. So the total area will be equal to the addition of each individual area.

$S=S_1+S_2$ (equation 2)

$S_1=a_1*h$

But we can write $S_1$ in terms of $\alpha _1$ , like this:

$\tan(\alpha _1)=\frac{h}{a_1} \\a_1=\frac{h}{\tan(\alpha _1)} \\S_1=\frac{h^{2} }{\tan(\alpha _1)}$

And for $S_2$ will be the same:

$S_2=\frac{h^{2} }{\tan(\alpha _2)}$

Replacing in the equation 2:

$S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})$

And replacing in the equation 1:

$h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})=a*h\\h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}$

So, we can replace all the known data in the last equation:

$h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}\\h=\frac{7220 ft}{(\frac{1 }{\tan(35.6)}+\frac{1}{\tan(58.2)})}\\h=3579,91 ft$

Then, the ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

6 0

2 years ago

Tulislah empt suku pertama dari barisan yang rumus suku ke n nya adalah Un = 3n^2-1

nirvana33 [79]

If n ∈ $\mathbb{N}$ then
$U_{0}=3(0)-1=-1$
$U_{1}=3(1)-1=2.$
$U_{2}=3(2^2)-1=11.$
$U_{3}=3(3^2)-1=26.$

6 0

3 years ago

Find the amount and compound interest on ₹ 5000 at 6% per annum, for 3 years, compounded annually.

stealth61 [152]

Given:

Principal = ₹ 5000

Compound rate of interest = 6% per annum

Time = 3 years

To find:

The amount after 3 years of compound interest.

Solution:

Formula for amount is:

$A=P\left(1+\dfrac{r}{100}\right)^t$

Where, P is principal, r is rate of interest in % and t is the number of years.

Putting P=5000, r=6 and t=3, we get

$A=5000\left(1+\dfrac{6}{100}\right)^3$

$A=5000\left(1+0.06\right)^3$

$A=5000\left(1.06\right)^3$

On further simplification, we get

$A=5000(1.018108216)$

$A=5000(1.191016)$

$A=5955.08$

Therefore, the amount after 3 years of compound interest is 5955.08.

4 0

3 years ago