All categories

2 years ago

Which of the following could NOT be the measures of the angles of a triangle?

1 answer:

8 0

the angles of a triangle always equals 180 degrees, so all of them cant be the measure of the angles of a triangle except for D

You might be interested in

Fred and wilma buy a home. they plan to make a down payment and carry a $90,000 mortgage. closing costs are $2,500 and are added

Korvikt [17]

$90,000 + $2,500 = $92,500

8 0

3 years ago

Read 2 more answers

(a) Jane pumped 54 gallons of water out of her pool. This was done over a period of 9 minutes at

STALIN [3.7K]

Answer: -6
Step-by-step explanation:
From the question, Elsa pumped 18 gallons of water out of her pool and this was done over a period of 3 minutes at a constant rate. Since it is done at a content rate, to get the change in the amount of water in the pool each minute, we divide 18 gallons by 3. This can be written mathematically as:
= 18 gallons ÷ 3
= 6 gallons
Change = -6
The change in the amount of water in the pool each minute will be -6gallons hope this helps

6 0

3 years ago

Read 2 more answers

Identify the expression with nonnegative limit values. More info on the pic. PLEASE HELP.

marshall27 [118]

Answer:

$\lim _{x\to 2}\:\frac{x-2}{x^2-2}\\\\ \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}\\\\ \lim _{x\to \frac{5}{2}}\left\frac{2x^2+x-15}{2x-5}\right$

Step-by-step explanation:

a) $\lim _{x\to 3}\:\frac{x^2-10x+21}{x^2+4x-21}=\lim \:_{x\to \:3}\:\frac{\left(x-7\right)\left(x-3\right)}{\left(x+7\right)\left(x-3\right)}=\lim \:_{x\to \:3}\:\frac{x-7}{x+7}=\frac{3-7}{3+7}=-\frac{4}{10}=-\frac{2}{5}$

b) $\lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=\lim \:_{x\to -\frac{3}{2}}\:\frac{\left(2x+3\right)\left(x-4\right)}{\left(2x+3\right)}=\lim \:\:_{x\to \:-\frac{3}{2}}\:\left(x-4\right)=-\frac{3}{2}-4\\ \\ \lim _{x\to -\frac{3}{2}}\left(\frac{2x^2-5x-12}{2x+3}\right)=-\frac{11}{2}$

c) $\lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0$

d) $\lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}$

e) $\lim _{x\to 3}\:\frac{x^2-8x+15}{x-3}=\lim \:_{x\to \:3}\:\frac{\left(x-3\right)\left(x-5\right)}{x-3}=\lim _{x\to 3}\left(x-5\right)=3-5=-2$

f) $\lim _{x\to \frac{5}{2}}\left(\frac{2x^2+x-15}{2x-5}\right)=\lim \:_{x\to \:\frac{5}{2}}\frac{\left(2x-5\right)\left(x+3\right)}{2x-5}=\lim \:\:_{x\to \:\:\frac{5}{2}}\left(x+3\right)=\frac{5}{2}+3=\frac{11}{2}$

4 0

3 years ago

F=5/4(n-90)<br> Find n <br> Helppp

padilas [110]

Answer:

n= 4f/5+90

Step-by-step explanation:

f= 5(n−90) /4 (simplify)

f * 4=5(n−90) (multiply 4 on both sides)

4f=5(n−90) (regroup)

4f/5 =n−90 (divide 5 on both sides)

4f/5 +90=n (add 90 to both sides)

4 0

3 years ago

The sides of a triangle are consecutive integers.if its perimeter is 126cm,find the lenghts of the sides.

Neko [114]

X + (x + 1) + (x + 2) = 126
3x = 123
x = 41
So the sides are 41, 42 and 43

3 0

3 years ago