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MrRissso [65]
3 years ago
9

Plz help me I will mark brainliest​

Mathematics
2 answers:
kozerog [31]3 years ago
7 0
The answer is Y= -4/3x+2
Igoryamba3 years ago
4 0

Answer:

(-15,0)(0,2)

\frac{2-0}{0+15} =\frac{2}{15} \\y=2/15x+b\\0=-2+b\\2=b\\

C. 2/15x+2

Step-by-step explanation:

You might be interested in
What is the cube root of 30
irakobra [83]
Cube root of 30:


Equivalent to 30^(1/3) = <span>3.10723250595
Approximately rounded to 3.12.</span>
3 0
3 years ago
Read 2 more answers
Cyrus, Ray and Billy visited the playground where their friends made a triangle out of string 9.5 m, 10.0 m, and 10.5 m long, an
Mkey [24]

Answer:

Billy next to the 65 degree angle

Ray next to the 60 degree angle

Cyrus next to the 55 degree angle

if we were to put this into A squared + B squared = C squared it would be Cyrus squared + Ray squared = Billy squared  

Also lol Billy Ray Cyrus (Miley Cyrus Dad)

Explaination:

It literally says it right here "Cyrus to stand by the 55o angle, Ray by the 60o angle, and Billy by the 65o angle. Determine which vertex each boy should stand next to."

3 0
4 years ago
The midpoint of AB is M(0,6). If the coordinates of A are (-3, 4), what are
Korolek [52]

Answer:

B(3,8)

Step-by-step explanation:

M(0,6)   A(-3,4)   B(x,y)

0=\frac{-3+x}{2}          6=\frac{4+y}{2}

0=-3+x         12=4+y

x=3               y=8

B(3,8)

3 0
4 years ago
Which expression is equivalent to 30m2+12m+18?? Please help ASAP !
ANTONII [103]

Answer:

c) 6(5m2+2m+3)

Step-by-step explanation:

Given expression = 30m2 + 12m + 18

The simplified form can be calculated by taking GCF

The GCF of 30m2, 12m and 18 is = 6

Therefore, the expression becomes

(6*5m2 + 6*2m + 6*3)

=> 6(5m2+2m+3) so option c is correct

7 0
4 years ago
Given: ΔABC is a right triangle. Prove: a2 + b2 = c2 The following two-column proof with missing justifications proves the Pytha
Anestetic [448]

Answer:

Transitive property of equality is not a justification for the proof.

Step-by-step explanation:

We draw a right angle ΔACB. CD is perpendicular to AB.

Let AC = a , BC = b , AB = c and CD = h

Now in ΔABC and ΔACD

∠C = ∠D and ∠A = ∠A

from AA similarity postulate

ΔABC  similar to ΔACD.

Hence,

           \frac{c}{a} = \frac{a}{x}

           a^{2} = c × x ·····················(1)

Now in ΔABC and ΔCBD

∠C = ∠D and ∠B = ∠B

from AA similarity postulates

ΔABC similar to ΔCBD

Hence,

           \frac{c}{b} = \frac{b}{y}

           b^{2} = c × y······················(2)

Add equation (1) and (2)

      a^{2} + b^{2} = cx + cy

      a^{2} + b^{2} = c(x+y)

      a^{2} + b^{2} = c^{2}               [because x+y=c]

Transitive property is not useful for this proof.

5 0
3 years ago
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