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Keith_Richards [23]
3 years ago
8

Hiiioo! Can someone please help me with this math question ❤️

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
4 0

Answer:

your answers will be

<E=25

M<F=65

M<G=90

Step-by-step explanation:

i hope it helps

have a nice day!!

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Two angles of a triangle have measures that total 135 degrees. Which equation can be used to find m, the measure of the third an
Mashutka [201]

Answer:Phytagorean thereom

Step-by-step explanation:it helps really try it.

8 0
3 years ago
A train travels 40 miles and 65 minutes to the nearest tenth of a mile how far does the train travel per minute
erik [133]

Answer:1.625

Step-by-step explanation:you just divide 65 and 40

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3 years ago
What is the x coordinate of the solution of the system? 2x+y=15 and x-y=3
Varvara68 [4.7K]

x - y = 3

x = 3 + y

x -3 = y

y = x-3

put it in 1st equation

2x +y = 15

2x + x - 3 = 15

3x - 3 = 15

3x = 15+3 = 18

x = 18/3 = 6

x = 6

6 0
3 years ago
∠F and ∠G are supplementary angles. If m∠F is two less than six times the measure of ∠G, find the measures of both angles
Mnenie [13.5K]

Answer:

∠ G = 26°, ∠ F = 154°

Step-by-step explanation:

Express ∠ F in terms of ∠ G, that is

F = 6G - 2

and since the angles are supplementary, then

G + 6G - 2 = 180, that is

7G - 2 = 180 ( add 2 to both sides )

7G = 182 ( divide both sides by 7 )

G = 26

Thus

∠ G = 26° and ∠ F = 6(26) - 2 = 156 - 2 = 154°

4 0
3 years ago
find the equations of the tangents to the curve y= x(x-1)(x+2) at the points where the curve cuts the x axis
antoniya [11.8K]

First of all, we compute the points of interest, i.e. the points where the curve cuts the x axis: since the expression is already factored, we have

x(x-1)(x+2) = 0 \iff x=0\ \lor\ x-1=0\ \lor\ x+2=0

Which means that the roots are

x=0\ \lor\ x=1\ \lor\ x=-2

Next, we can expand the function definition:

y = x(x-1)(x+2) = x^3 + x^2 - 2x

In this form, it is much easier to compute the derivative:

y' = 3x^2+2x-2

If we evaluate the derivative in the points of interest, we have

y'(-2) = 6,\quad y'(0)=-2,\quad y'(1)=3

This means that we are looking for the equations of three lines, of which we know a point and the slope. The equation

y-y_0=m(x-x_0)

is what we need. The three lines are:

y-0=6(x+2) \iff y = 6x+12  This is the tangent at x = -2

y-0=-2(x-0) \iff y = -2x  This is the tangent at x = 0

y-0=3(x-1) \iff y = 3x-3  This is the tangent at x = 1

7 0
3 years ago
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