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2 years ago

Angles y and z are formed by two intersecting lines. The measure of z is 146°. What is measure of y?

Mathematics

1 answer:

AlladinOne [14]2 years ago

3 0

Answer: 34

Step-by-step explanation: A line is 180 degrees, so just subract the value that you already have from 180 and you'll have your answer.

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What is the awnser to the problem b-9.5<7

OlgaM077 [116]

Answer:

b < 16.5

Step-by-step explanation:

b - 9.5 + 9.5 < 7 + 9.5

b < 16.5

hope this helped :D

5 0

2 years ago

Solve the equation. \dfrac{11}6 =n+\dfrac79 6 11 =n+ 9 7 start fraction, 11, divided by, 6, end fraction, equals, n, plus, s

kvasek [131]

11/6 = n + 7/9
11/6 - 7/9 = n = 19/18

5 0

3 years ago

In order for the parallelogram to be a<br> rhombus, x = [?].<br> (5x + 25)<br> (12x + 11)

LiRa [457]

A parallelogram is also a rhombus if the diagonal is a bisector of an angle enclosed by the two adjacent sides of a parallelogram.

In our case it means,

$5x+25=12x+11$

$7x=14\implies x=\boxed{2}$

Hope this helps.

8 0

3 years ago

ANSWER FAST!

salantis [7]

Answer:C

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

2 years ago