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3 years ago

7y + 3y = (I’m just not good at algebra I try but can someone help)

Mathematics

2 answers:

jarptica [38.1K]3 years ago

4 0

Isnt this 10y? 7y+3y=10

zhannawk [14.2K]3 years ago

3 0

7y+3y= 10y (combine like terms)

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Do 5/7 and 30/42 form a proportion

GREYUIT [131]

5/7
6*5/6*7
30/42

30/42 = 30/42

Yes, 5/7 and 30/42 form a proportion.

4 0

3 years ago

Find the slope(6,8) and (9,10)

USPshnik [31]

name the points

a=(x1,y1) b=(x2,y2)

a=(6,8) b=(9,10)

use the slope formula

$m=\frac{y2-y1}{x2-x1}$

replace

$\begin{gathered} m=\frac{10-8}{9-6} \\ m=\frac{2}{3} \end{gathered}$

answer= The slope is equal to 2/3

a=(9,10) b=(6,8)

using the formula

$\begin{gathered} m=\frac{y2-y1}{x2-x1} \\ m=\frac{8-10}{6-9} \\ m=\frac{-2}{-3}=\frac{2}{3} \end{gathered}$

slope will also be 2/3

5 0

1 year ago

Determine whether the number is rational or irrational and round it to the nearest hundredth. - 3/7

ivann1987 [24]

Answer:

Yes, it's a rational number

Step-by-step explanation:

-3/7. It is a rational because. It can be written as p/q form. Where p/q are non zero number. And in decimal it would be -0.42

6 0

3 years ago

Read 2 more answers

John started a home delivery service from several local restaurants. He charges a $5 tip on the delivery, plus $0.75 for every m

padilas [110]

Answer:c(d) = 5 + (0.75*d)

Step-by-step :

In this equation the cost (c) of the distance traveled (d) has a base rate of 5, and increases by .75 for every mile traveled (d)

6 0

2 years ago

Read 2 more answers

What is the range of the equation

a_sh-v [17]

The range of the equation is $y>2$

Explanation:

The given equation is $y=2(4)^{x+3}+2$

We need to determine the range of the equation.

<u>Range:</u>

The range of the function is the set of all dependent y - values for which the function is well defined.

Let us simplify the equation.

Thus, we have;

$y=2 \cdot 4^{x+3}+2$

This can be written as $y=2^{1+2(x+3)}+2$

Now, we shall determine the range.

Let us interchange the variables x and y.

Thus, we have;

$x=2^{1+2(y+3)}+2$

Solving for y, we get;

$x-2=2^{1+2(y+3)}$

Applying the log rule, if f(x) = g(x) then $\ln (f(x))=\ln (g(x))$ , then, we get;

$\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)$

Simplifying, we get;

$(1+2(y+3)) \ln (2)=\ln (x-2)$

Dividing both sides by $\ln (2)$ , we have;

$2 y+7=\frac{\ln (x-2)}{\ln (2)}$

Subtracting 7 from both sides of the equation, we have;

$2 y=\frac{\ln (x-2)}{\ln (2)}-7$

Dividing both sides by 2, we get;

$y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}$

Let us find the positive values for logs.

Thus, we have,;

$x-2>0$

$x>2$

The function domain is $x>2$

By combining the intervals, the range becomes $y>2$

Hence, the range of the equation is $y>2$

7 0

3 years ago