All categories

2 years ago

Write an equation of the line that passes through (–5, 1) and (7, 1)

Mathematics

1 answer:

monitta2 years ago

6 0

I think it’d be y = x + 1

You might be interested in

write a symbolic rule to represent the relationship in which the output variable y is 12 less than the input variable x.what is

sveticcg [70]

Answer:

the value of output is 29

Step-by-step explanation:

the output variable y is 12 less then the input variable x.

Input variable is x, output variable is y

output values = input variable - 12

now plug in the variable in the equation

output values = input variable - 12

Given the input value is 41

Substitute 41 for x

the value of output is 29

4 0

2 years ago

WILL GIVE BRAINLEST plz help mee

LUCKY_DIMON [66]

Answer:

-2 3/4

Step-by-step explanation:

i gotchu

8 0

3 years ago

Read 2 more answers

What is the image of (3,9) after a dilation by a scale factor of 3 centered at the origin

Sav [38]

Answer:

(9, 27 )

Step-by-step explanation:

Since the dilatation is centred at the origin, multiply each of the coordinates of the original point by the scale factor of 3

(3, 9 ) → (3(3), 3(9)) → (9, 27 )

7 0

3 years ago

Read 2 more answers

Need help ASAP

joja [24]

Part (1) : The solution is $729$

Part (2): The solution is $\frac{1}{16 x^{8}}$

Part (3): The solution is $\frac{2 x^{2}}{3 y z^{7}}$

Explanation:

Part (1): The expression is $3^{2} \cdot3^{4}$

Applying the exponent rule, $a^{b} \cdot a^{c}=a^{b+c}$ , we get,

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

Adding the exponent, we get,

$3^{2} \cdot3^{4}=3^6=729$

Thus, the simplified value of the expression is $729$

Part (2): The expression is $\left(2 x^{2}\right)^{-4}$

Applying the exponent rule, $a^{-b}=\frac{1}{a^{b}}$ , we have,

$\left(2 x^{2}\right)^{-4}=\frac{1}{\left(2 x^{2}\right)^{4}}$

Simplifying the expression, we have,

$\frac{1}{2^4x^8}$

Thus, we have,

$\frac{1}{16 x^{8}}$

Thus, the value of the expression is $\frac{1}{16 x^{8}}$

Part (3): The expression is $\frac{2 x^{4} y^{-4} z^{-3}}{3 x^{2} y^{-3} z^{4}}$

Applying the exponent rule, $\frac{x^{a}}{x^{b}}=x^{a-b}$ , we have,

$\frac{2x^{4-2}y^{-4+3}z^{-3-4}}{3}$

Adding the powers, we get,

$\frac{2x^{2}y^{-1}z^{-7}}{3}$

Applying the exponent rule, $a^{-b}=\frac{1}{a^{b}}$ , we have,

$\frac{2 x^{2}}{3 y z^{7}}$

Thus, the value of the expression is $\frac{2 x^{2}}{3 y z^{7}}$

8 0

3 years ago

What is the measure of Arc RS?

IRISSAK [1]

Answer:

4.54 inches

Step-by-step explanation:

3 0

2 years ago