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aniked [119]
2 years ago
10

Which expression is equivalent to 2(5m) + m? a) 11m b) 12m c) 5m + 2 d) 7m + 2m

Mathematics
1 answer:
Aliun [14]2 years ago
6 0

Answer:

choice a.)  11m

Step-by-step explanation:

2(5m) + m  =  10m + m = 11m

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Gcf of 6x^3, 8x^5, 10x^7
Art [367]

Answer: 2x^3

Step-by-step explanation: Find the prime factors of each term in order to find the GCF (Greatest Common Denominator).

Hope this helps you out! ☺

5 0
2 years ago
HELP PLEASE AND THANK YOU 1JAAB
Dafna1 [17]

Answer:

\bold{Q1}\ side\ BD\\\\\bold{Q2}\ \angle5\ \leftrightarrow\ \angle6\\\\\bold{Q3}\ \triangle TAP\\\\\bold{Q4}\ \triangle APT

Step-by-step explanation:

For Q1 and Q2

\text{If}\ AB\cong AC,\ BD\cong CD,\ RSTU\ \text{is a parallelogram (rectangle)},\ \text{then}\\\\\angle 1\ \leftrightarrow\ \angle2\\\angle3\ \leftrightarrow\ \angle4\\\angle C\leftrightarrow\ \angle B\\\angle 5\ \leftrightarrow\ \angle6\\\angle7\ \leftrightarrow\ \angle 8\\\angle R\ \leftrightarrow\ \angle T\\\angle S\ \leftrightarrow\ \angle U

\overline{AB}\ \leftrightarrow\ \overline{AC}\\\overline{BD}\ \leftrightarrow\ \overline{CD}\\\\\overline{RS}\ \leftrightarrow\ \overline{TU}\\\overline{ST}\ \leftrightarrow\ \overline{UR}

For Q3 and Q4

O\ \leftrightarrow\ T\\B\ \leftrightarrow\ A\\R\ \leftrightarrow\ P\\\\E\ \leftrightarrow\ X\\F\ \leftrightarrow\ W\\D\ \leftrightarrow\ T

4 0
3 years ago
Points A and B are graphed points on the coordinate plane below. Which ordered pair is the coordinate location which is halfway
Setler [38]

Answer:

I cant see the image it doesnt seem to be loading but here is how to do the problem-->.

Add both the x coordinates and divide by 2 (this is the new x coordinate for the midpt)

next:

Add both y coordinates and divide by 2(this is the new y coordinate for the midpt)

Step-by-step explanation:

5 0
2 years ago
Write an equation of the hyperbola given that the center is at (2, -3), the vertices are at (2, 3) and (2, - 9), and the foci ar
zavuch27 [327]
Check the picture below.

so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).

because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,

\bf \textit{hyperbolas, vertical traverse axis }
\\\\
\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1
\qquad 
\begin{cases}
center\ ( h, k)\\
vertices\ ( h,  k\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2 + b ^2}\\
asymptotes\quad  y= k\pm \cfrac{a}{b}(x- h)
\end{cases}\\\\
-------------------------------

\bf \begin{cases}
h=2\\
k=-3\\
a=6\\
c=2\sqrt{10}
\end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1
\\\\\\
\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1
\\\\\\
c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2
\\\\\\
4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2
\\\\\\
\sqrt{4}=b\implies 2=b\\\\
-------------------------------\\\\
\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1

3 0
2 years ago
Why are expert verified people so extra like they put too much information
lys-0071 [83]

Answer:

to make sure you understand the question

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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