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3 years ago

OF bisects EOF. EOF=y+30 and FOG=3y-50. solve for y.

1 answer:

3 0

It always helps to draw a picture. Given the information, Segment OF is the center line that is bisecting this angle.

Since it's bisecting (cutting in half)... we can simply set the two angles equal to each other.

y+30=3y-50
-2y+30=-50
-2y=-80
y = 40

C)40.

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What are the domain and range of the function f(x)=(√x-7)+9?

Yanka [14]

Answer:

the domain of the function f(x) is $x\ge 7$

the range of the function f(x) is $f(x)\ge 9$

Step-by-step explanation:

Consider the parent function $y=\sqrt{x}.$

The domain og this function is $x\ge 0,$ the range of this function is $y\ge 0.$

The function $f(x)=\sqrt{x-7}+9$ is translated function $y=\sqrt{x}$ 7 units to the right and 9 units up, so

the domain of the function f(x) is $x\ge 7$

the range of the function f(x) is $f(x)\ge 9$

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4 years ago

Write an equation of the line with the given slope and y-intercept.

Snezhnost [94]

Answer:

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

Step-by-step explanation:

general slope intercept equation of a straight line is given as

$y=mx+b$

where m is slope and b is y-intercept

here

$m=\frac{4}{3}$

and

$b=0$

8 0

3 years ago

If 3x+2=11<br> then x= ?

leonid [27]

Answer:

X=3

Step-by-step explanation:

What you have to do is minus 2 from 11 and you get 9

Now divide 3 into 9 and you get 3

x=3 you can also plug 3 for x to see if the equation is right

8 0

3 years ago

Read 2 more answers

Given ∆QRS≅∆TUV, QS=4v+3, and TV=7v-9, find the length of QS and TV.?

Anon25 [30]

If we are supposed to assume that QS=TV

4v+3=7v-9
minus 4v both sides
3=3v-9
add 9
12=3v
divide 3
4=v
v=4

sub back

4v+3=QS=TV
4(4)+3=QS=TV
16+3=QS
19=QS=TV

then answer is C

7 0

3 years ago

Read 2 more answers

Kindly answer this one. Thank you.

steposvetlana [31]

Answer:

See below.

Step-by-step explanation:

I'm assuming these questions are about the Midline Theorem (segment AL joins the midpoints of the non-parallel sides.

♦ The midline's length is the average of the lengths of the top and bottom parallel sides.

$AL=\frac{OR+CE}{2}$