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2 years ago

Which of the following is a true statement about the triangles shown on the graph

Mathematics

1 answer:

sveticcg [70]2 years ago

5 0

It's D. Slopes are the same, because if you divide opposite and adjacent side of both triangles, you get the same result. Count the squares, for the smaller triangle it's 2/2=1 and for the bigger one it's 4/4=1.

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Please help me on this

dlinn [17]

Answer:

-q

Step-by-step explanation:

We know that we are trying to find an expression equivelant to p - q:

p - q can be p - 1q

We can then put parenthesis around -1q:

p + (-1q)

We then can simplify:

p + (-q)

5 0

2 years ago

Angie got some money for his birthday. He spent 1/5 of it on dog treats for his puppy. Then he divided the remainder equally amo

prohojiy [21]

Answer:

225

Step-by-step explanation:

60 x 3 = 180 ($60 for each charity and there's 3 charities)

180 / (4/5) = 225 (if $180 is four fifths of his total just divide 180 by 4/5 to get the answer)

3 0

3 years ago

10 is 8% of a, and 5 is 4% of b. c equals b/a. What is the value of c?? step by step explanation pls...

klio [65]

Answer:

c = 1

Step-by-step explanation:

"10 is 8% of a": 0.08a = 10, or a = (10/0.08)

"5 is 4% of b": 0.04b = 5, or b = (5/0.04)

"c equals b/a": c = (b/a)

----

What is the value of c?

c = (b/a)

c = ((5/0.04)/(10/0.08))

c = (5/0.04)*(0.08/10)

c = (5/10)*(0.08/0.04)

c = (1/2)*(2)

c = 1

5 0

2 years ago

(x +y)^5 Complete the polynomial operation

Vesna [10]

Answer:

Please check the explanation!

Step-by-step explanation:

Given the polynomial

$\left(x+y\right)^5$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=x,\:\:b=y$

$=\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$

so expanding summation

$=\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$

solving

$\frac{5!}{0!\left(5-0\right)!}x^5y^0$

$=1\cdot \frac{5!}{0!\left(5-0\right)!}x^5$

$=1\cdot \:1\cdot \:x^5$

$=x^5$

also solving

$=\frac{5!}{1!\left(5-1\right)!}x^4y$

$=\frac{5}{1!}x^4y$

$=\frac{5x^4y}{1}$

$=5x^4y$

similarly, the result of the remaining terms can be solved such as

$\frac{5!}{2!\left(5-2\right)!}x^3y^2=10x^3y^2$

$\frac{5!}{3!\left(5-3\right)!}x^2y^3=10x^2y^3$

$\frac{5!}{4!\left(5-4\right)!}x^1y^4=5xy^4$

$\frac{5!}{5!\left(5-5\right)!}x^0y^5=y^5$

so substituting all the solved results in the expression

$=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

Therefore,

$\left(x\:+y\right)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

6 0

2 years ago

Please help ASAP! Find the midpoint between the complex numbers.

Ivenika [448]

$\bf \begin{cases}
z1=9-9i\implies &(9~,~-9)\\
z2=10-9i\implies &(10~,~-9)
\end{cases}$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 9 &,& -9~) 
% (c,d)
&&(~ 10 &,& -9~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right)
\\\\\\
\left( \cfrac{10+9}{2}~~,~~\cfrac{-9-9}{2} \right)\implies \left( \cfrac{19}{2}~~~,~~\cfrac{-18}{2} \right)
\\\\\\
\left( \cfrac{19}{2}~~~,~~-9 \right)\implies \left( 9\frac{1}{2}~~~,~~-9 \right)$

3 0

3 years ago