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3 years ago

Write an equation of a line in slope-intercept from with the given slope, slope:–3/4, y-intercept: –2

Mathematics

1 answer:

larisa [96]3 years ago

7 0

Answer: y=-3/4x-2

Step-by-step explanation:

Write an equation of a line in slope-intercept from with the given slope, slope:–3/4, y-intercept: –2

The y-intercept is at (0,-2)

Then we write the equation using y=mx+b becuase that is the point slope form.

m=-3/4 which was given in the question

because y-intercept is at (0,-2)

y=-2

x=0

-2=-3/4(0)+b

-2=0+b

-2=b

now we know that m is -3/4 and b is -2

Hence the equation is y=-3/4x-2

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What is nine plus ten? If you get it right I’ll give you brainliest

Finger [1]

Answer:

Step-by-step explanation:

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3 years ago

Given: △ACM, m∠C=90°, CP ⊥ AM

larisa86 [58]

Answer: The answer is $3\dfrac{4}{7}.$

Step-by-step explanation: Given in the question that ΔAM is a right-angled triangle, where ∠C = 90°, CP ⊥ AM, AC : CM = 3 : 4 and MP - AP = 1. We are to find AM.

Let, AC = 3x and CM = 4x.

In the right-angled triangle ACM, we have

$AM^2=AC^2+CM^2=(3x)^2+(4x)^2=9x^2+16x^2=25x^2\\\\\Rightarrow AM=5x.$

Now,

$AM=AP+PM=AP+(AP+1)\\\\\Rightarrow 2AP=AM-1\\\\\Rightarrow 2AP=5x-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A)$

Now, since CP ⊥ AM, so ΔACP and ΔMCP are both right-angled triangles.

So,

$CP^2=AC^2-AP^2=CM^2-MP^2\\\\\Rightarrow (3x)^2-AP^2=(4x)^2-(AP+1)^2\\\\\Rightarrow 9x^2-AP^2=16x^2-AP^2-2AP-1\\\\\Rightarrow 2AP=7x^2-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(B)$

Comparing equations (A) and (B), we have

$5x-1=7x^2-1\\\\\Rightarrow 5x=7x^2\\\\\Rightarrow x=\dfrac{5}{7},~\textup{since }x\neq 0.$

Thus,

$AM=5\times\dfrac{5}{7}=\dfrac{25}{7}=3\dfrac{4}{7}.$

8 0

3 years ago

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jekas [21]

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3 years ago

Last week Julia ran 30 miles more than Pranav.Julia ran 47 miles.How many miles did Pranav run?

andrew-mc [135]

17 miles as it is 47-30

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3 years ago

A cyclist travels at distance of 400 meters in 120 seconds towards school, calculate his speed. (Show your work)

DochEvi [55]

we know that

The scalar magnitude of the velocity vector is the speed. The speed is equal to

$Speed=\frac{distance}{time}$