All categories

3 years ago

What’s the terms and coefficients of 7p-3r+6-5s

Mathematics

2 answers:

Maslowich3 years ago

8 0

Answer:

just Move the 6

7p−3r−5s+6

Step-by-step explanation:

rosijanka [135]3 years ago

5 0

Answer:

See below in bold.

Step-by-step explanation:

7p - 3r + 6 - 5s.

The terms are

7p, 3r, 6 and -5s.

The coefficients are

7, -3 and -5.

(6 is a constant).

You might be interested in

Determine the domain of the function (fog) (x) where f(x) = 3x-1/x-4 and g(x)=x+1/x

Mrac [35]

Answer:

$(-\infty,0)\cup(0, \frac{1}{3})\cup (\frac{1}{3},\infty)$

Step-by-step explanation:

The given functions are $f(x)=\frac{3x-1}{x-4}$ and $g(x)=\frac{x+1}{x}$ .

We now composed the two functions to find:

$(f\circ g)(x)=f(g(x))$

$\implies (f\circ g)(x)=f(\frac{x+1}{x})$

$\implies (f\circ g)(x)=\frac{3(\frac{x+1}{x})+1}{\frac{x+1}{x}-4}$

$\implies (f\circ g)(x)=\frac{4x+3}{1-3x}$

This function is defined if the denominator is not zero.

$1-3x\ne0$

$x\ne\frac{1}{3}$

We write this in interval notation as:

$(-\infty,\frac{1}{3})\cup (\frac{1}{3},\infty)$

We need to be cautious here as x=0 is not in the domain of g(x).

Therefore the domain of $(f\circ g)(x)$ is

$(-\infty,0)\cup(0, \frac{1}{3})\cup (\frac{1}{3},\infty)$

4 0

3 years ago

What is 10000 more than 46952 and what is the missing<br> number

Morgarella [4.7K]

10000 more than 46952 is 56952

3 0

3 years ago

Read 2 more answers

Find the difference. Check your answer by adding.<br><br> 652,203 - 62,285

Anna11 [10]

The answer is 589,918
Take 652203 and subtract 62,285 from that and you get 589,918

7 0

2 years ago

Given: JK tangent, KH=16, HE=12 Find: JK.

Y_Kistochka [10]

Answer: $JK=8$

Step-by-step explanation:

You can observe in the figure that JK is a tangent and KH is a secant and both intersect at the point K. Then, according to the Intersecting secant-tangent Theorem:

$JK^2=KE*KH$

You know that:

$KH=KE+HE$

Then KE is:

$KE=KH-HE$

$KE=16-12$

$KE=4$

Now you can substitute the value of KE and the value of KH into $JK^2=KE*KH$ and solve for JK. Then the result is:

$JK^2=4*16\\JK^2=64\\JK=\sqrt{64}\\JK=8$

7 0

3 years ago

What’s the volume of a triangular prism?

kati45 [8]

Answer:

$V = 0.5 X b X a X h$

4 0

2 years ago