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2 years ago

Hellpppppp please I will give brainliest

Mathematics

2 answers:

liberstina [14]2 years ago

8 0

Answer:

y = 5/4x - 1

Step-by-step explanation:

if you look at the graph you will probably be able to tell

im not trying to be me if you think

Vedmedyk [2.9K]2 years ago

4 0

Answer:

ITS THE 2ND ONE

Step-by-step explanation:

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What is the Ratio of 18:24, 4:70, 35:105, 22:7, 12:165, In fractions in their Simplest form?

Degger [83]

18:24 = 3/4
4:70 = 2/35
35:105 = 1/3
22:7 = 22/7
12:165 = 4/55

7 0

3 years ago

Combine like terms to write an equivalent expression. Then use y = 5 to prove the expressions are equivalent. 1 + y + y + y + 1

NeX [460]

Answer:

The equivalent expression is

✔ 3y + 2

The value of both expressions when y = 5 is

✔ 17

The expressions are

✔ equivalent

Step-by-step explanation:

3 0

2 years ago

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Which description best compared the graph of the two functions below?

Alenkinab [10]

Answer:

The description that best compares the graph of the two function is:

The line for function A is steeper.

Step-by-step explanation:

5 0

2 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

2 years ago

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PLEEEAAAAAASE HELP! What is the answer to this question?

IrinaVladis [17]

Answer:

23 units

Step-by-step explanation:

Perimeter is the sum of side lengths

Let's find the sides first

AB = √(-2- 2)² + (-4+1)² = 5
BC = (2 - (-1)) = 3
CD = √(-1- 2)²+(5-2)² = √18 = 3√2 = 4.24
DE = √(-4 -(-1))²+(2-5)²= √18 = 3√2 = 4.24
AE = √(-4-(-2)²+(2-(-4))²=√40= 2√10 = 6.32

Perimeter is

P = 5 + 3 + 4.24 + 4.24 + 6.32 = 22.8 ≈ 23 rounded to the nearest whole number

7 0

3 years ago

Hellpppppp please I will give brainliest​

Hellpppppp please I will give brainliest