All categories

3 years ago

Determine whether the two triangles are similar. PLEASE HELP!!

Mathematics

1 answer:

zvonat [6]3 years ago

6 0

I think that it is C: PQR~XY by AA because there are only two angles that you are given ;the boxes represent 90, so each box has 90 and 62.

You might be interested in

Solve this quadratic equation. x^2+2x-22=0

torisob [31]

X=⁻3 and x=1
because
a+b=2 which means a=3 and b=⁻1 which added up equals 2
a×b=⁻22 which means that 11 which is a×⁻2 which is b=⁻22

7 0

3 years ago

Is this correct?- <br><br> Also please don’t just say “yes” or “no”-

hichkok12 [17]

Answer:

a) No

b) No

c) No

d) Yes

3 0

3 years ago

18 2/5 - n = -10 1/3

Wewaii [24]

Answer:

The answer is $n=28\frac{11}{15}$ .

Step-by-step explanation:

Given:

18 2/5 - n = -10 1/3.

Now, to solve the equation:

$18\frac{2}{5} - n = -10 \frac{1}{3}$

Converting the mixed fractions to improper:

⇒ $\frac{92}{5} -n=-\frac{31}{3}$

Moving the numbers on one side and variable to the other:

⇒ $\frac{92}{5} +\frac{31}{3}=n$

Now, solving the factions:

⇒ $\frac{92\times 3+31\times 5}{15}=n$

⇒ $\frac{276+155}{15}=n$

⇒ $\frac{431}{15}=n$

⇒ $28\frac{11}{15}=n$

Therefore, the answer is $n=28\frac{11}{15}$ .

3 0

3 years ago

PLEASE HELP ME WITH THIS PROBLEM IVE BEEN ON IT FOR 3 DAYS NOW!

Len [333]

85 x 3 = 255
first 2 test grades were 81 and 86:
81 + 86 = 167

Need to find the 3rd to have exact average 85:
255 - 167 = 88

So her 3rd test needs to make 88 to make average of 85

Double check: 81 + 86 + 88 = 255 = 85 + 85 + 85

3 0

3 years ago

Prove 2√(x) + 1/√(x + 1) <= 2√(x+1) for all x in [0,inf)

EleoNora [17]

Start by multiplying each side of the inequality by $\sqrt{x + 1}$ and simplifying:

$2 \sqrt x + \frac{1}{\sqrt{x+1}} \leq 2 \sqrt{x + 1}$
$(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})$
$2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)$
$2 \sqrt{x^2 + x} + 1 \leq 2x + 2$
$2 \sqrt{x^2 + x} \leq 2x + 1$
$\sqrt{x^2 + x} \leq x + \frac{1}{2}$

From here, we can square both sides to get

$x^2 + x \leq (x + \frac{1}{2})^2$
$x^2 + x \leq x^2 + x + \frac{1}{4}$
$0 \leq \frac{1}{4}$ , which is always true.

3 0

2 years ago