How many factors in the expression 8(+4)(y+4)(z²+4z+7)have exactly two terms?

1 answer:

6 0

In mathematics, a term is defined as a number or a variable, or a combination of both multiplied together.

Therefore, in the expression above, there are 4 factors, which consist of terms.
However, the only factor in which there is two terms is (y + 4), because both y and 4 are separate terms.

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Please help me. Need it.

Lunna [17]

Hey There!
win 10 lose 12
win 15 lose 18

5×2=10, 6×2=12
5×3=15, 6×3=18

Hope This Helps!!!

4 0

3 years ago

Look at the following numbers:

jok3333 [9.3K]

Answer:

The correct answer is: -7, 7

Step-by-step explanation:

Given numbers are:

-7, 0, 7, 14

We have to check all pairs of numbers with respect to sum to check that which pair of numbers sums up to zero. So first of all, we will take -7 and will find its sum with other numbers.

So,

$-7+0 = -7\\-7+7=0\\-7+14 = 7$

We can see that the sum of -7 and 7 is zero.

Hence,

The correct answer is: -7, 7

8 0

3 years ago

Read 2 more answers

Use the laplace transform to solve the given initial-value problem. y' 5y = e4t, y(0) = 2

Basile [38]

The Laplace transform of the given initial-value problem

$y' 5y = e^{4t}, y(0) = 2$ is mathematically given as

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is mathematically given as

$&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}$

$\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}$

$put $s=-s \Rightarrow a_{1}=\frac{17}{9}$$

$\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}$

In conclusion, Taking inverse Laplace tranoform

$L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\$

$y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}$