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2 years ago

Is a number multiplied by a variable

Mathematics

2 answers:

Bas_tet [7]2 years ago

6 0

A number multiplied by a variable is a coefficient

Anna35 [415]2 years ago

3 0

Answer:

imma have to answer yes.

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a hawk soars at an altitude of 1,800ft. if the awk descends to the ground in 45 min, what is its vertical speed?

Verizon [17]

Does it come with any anwsers

4 0

2 years ago

Suppose △ITH≅△APG.<br><br> Which congruency statement is true?

kupik [55]

It is given that the triangles ITH and APG are congruent.

When the two triangles are congruent, then the corresponding angles and sides are congruent to each other.

By being congruent, the corresponding angles which are congruent are:

$IT \cong AP, TH \cong PG, IH \cong AG$

Therefore, $IH \cong AG$

SO, Option 2 is the correct answer.

5 0

2 years ago

Read 2 more answers

Exponents: Can me help me? This is the one, I just can't get.

aliina [53]

Answer:

Step-by-step explanation:

x^y * x^-y

When the bases are the same, we add the exponents when multiplying

x^(y-y)

x^0

Anything to the zero power is 1 (except 0)

4 0

2 years ago

PLEASE HELP ME!!

Step2247 [10]

Answer:

A = $10,441.68

A = P + I where

P (principal) = $10,400.00

I (interest) = $41.68

Step-by-step explanation:

First, convert R as a percent to r as a decimal

r = R/100

r = 0.02/100

r = 0.0002 rate per year,

Then solve the equation for A

A = P(1 + r/n)nt

A = 10,400.00(1 + 0.0002/2)(2)(20)

A = 10,400.00(1 + 0.0001)(40)

A = $10,441.68

Summary:

The total amount accrued, principal plus interest, with compound interest on a principal of $10,400.00 at a rate of 0.02% per year compounded 2 times per year over 20 years is $10,441.68.

8 0

2 years ago

Use matrix addition to solve this equation: B + 15 −7 4 0 1 2 = 1 2 12 4 0 2 b11 = b12 = b13 = b21 = 4 b22 = −1 b23 = 0

Arada [10]

Answer:

$b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0$

Step-by-step explanation:

The given matrix addition is

$B+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$

We need to find the elements of matrix B.

Let $B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}$

Substitute the value of matrix.

$\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$