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klio [65]
3 years ago
10

What equations represents the line that is parallel to 3x-4y=7 and passes through the point (-4,-2) check all that apply

Mathematics
1 answer:
valina [46]3 years ago
6 0

Answer:

y=\frac{3}{4}x + 3

Step-by-step explanation:

Parallel lines have the same slope. To find the slope, convert 3x-4y=7 into y=mx+b form.

3x - 4y = 7

-4y = 7 -3x

y= -7/4 + 3/4 x

The slope of the line is 3/4. Using this slope, substitute m=3/4 and the point (-4,-2) into the point slope form.

y-y_1=m(x-x_1)\\y--2=\frac{3}{4}(x--4)\\y+2=\frac{3}{4}(x+4)\\y=\frac{3}{4}x + 3

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I actually don’t get this and I have a test in 2 days please help what is g(x) and how
nataly862011 [7]

first off, is noteworthy that's the graph of an exponential function, thus the function will be along the lines of g(x) = abˣ , now, what's "a" and "b" values?

well, let's take a peek when x = 0 and x = 1.

\bf g(x) = ab^x \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x = 0\\ y = 1 \end{cases}\implies 1=ab^0\implies 1=a(1)\implies \boxed{1=a} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x = 1\\ y = 4 \end{cases}\implies 4 = ab^1\implies 4=1b^1\implies \boxed{4=b} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill g(x) = 4^x\qquad \qquad \qquad \begin{array}{|c|c|ll} \cline{1-2} x&y\\ \cline{1-2} -2&\frac{1}{4^2}\to \frac{1}{16}\\ -1&\frac{1}{4}\\ 0&1\\ 1&4\\ 2&16\\ \cline{1-2} \end{array}~\hfill

8 0
3 years ago
Please answer asap...
Anon25 [30]

Answer:

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To get this answer all you need to do is

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7 0
3 years ago
Read 2 more answers
Needs answers <br> Will mark brainliest
Elden [556K]

QUESTION 1:

Required angles are: m<2=75°, m<3=75°, m<4= 105°, m<5= 105°, m<6=75°, m<7=75°, m<8=105°

QUESTION 2:

Required angles are: m<1=100° m<2=80°, m<4= 100°, m<5= 100°, m<6=80°, m<7=80°, m<8=100°

Step-by-step explanation:

Question 1:

if m<1 =105°

We need to find the remaining angles.

The corresponding angles of a traversal are same. so the corresponding angle of <1 is angle 5 So, m<5 = 105°

The vertical angles are also equal. so, vertical angle of <1 is <4. So, m<4 = 105°

Vertical angles of m<5 is m<8 so, m<8=105°

m<1 and m<2 are supplementary angles of each other so their sum will be equal to 180°

So, m<1+m<2=180

105+m<2=180

m<2=75°

Same goes for m<3 and m<4 so, m<3 = 75°

m<5 and m<6 so, m<6 = 75°

m<7 and m<8 so, m<7= 75°

So, required angles are: m<2=75°, m<3=75°, m<4= 105°, m<5= 105°, m<6=75°, m<7=75°, m<8=105°

Question No 2

if m<3=80°

We need to find the remaining angles.

The corresponding angles of a traversal are same. so the corresponding angle of <3 is angle 7 So, m<7 = 80°

The vertical angles are also equal. so, vertical angle of <3 is <2. So, m<2 = 80°

The Vertical angle of <7 is <6 so, m<6= 80°.

m<1 and m<2 are supplementary angles of each other so their sum will be equal to 180°

So, m<1+m<2=180

m<1+80=180

m<1=100°

Same goes for m<3 and m<4 so, m<4 = 100°

m<5 and m<6 so, m<5 = 100°

m<7 and m<8 so, m<8= 100°

So, required angles are: m<1=100° m<2=80°, m<4= 100°, m<5= 100°, m<6=80°, m<7=80°, m<8=100°

Keywords: Traversal of parallel lines

Learn more about traversal of parallel lines at:

  • brainly.com/question/1598227
  • brainly.com/question/3227215
  • brainly.com/question/10483199

#learnwithBrainly

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The answer to the math problem is 8y^12

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