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a_sh-v [17]
3 years ago
7

Solve for x. 4(x + 3) = x +42 x = [?]

Mathematics
1 answer:
laila [671]3 years ago
7 0

Answer: x=10

Step-by-step explanation:

4(x+3)=x+42\\4x+12=x+42\\4x-x=42-12\\3x=30\\x=10

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Please help!! (Number 73)
Rasek [7]
So hmm notice the picture below

one is 7x5x1.75 the volume of a rectangular prism is V = length * width * height

so 7*5*1.75 gives us 61.25 ft³

the second one, is larger by some width and length we dunno, but we know that it required that 61.25 plus an extra 17.5 to fill it up, so its volume is 61.25 + 17.5 or 78.75

the height is the same... so \bf V=l\cdot w\cdot h\qquad 
\begin{cases}
l=length\\
w=width\\
h=height\\ --------\\
h=1.75\\
V=78.75
\end{cases}\implies 78.75=lw\cdot 1.75
\\\\\\
\cfrac{78.75}{1.75}=lw\implies 45=lw

so.. if you factor 45, to two factors, one will be the length, the other the width

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2 years ago
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77julia77 [94]

Answer:

D

Explanation:

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4 0
2 years ago
If quadrilateral PQRS is a rectangle, then which of the following is true? ∠PSQ ≅ ∠QSR segment SR ≅ segment RQ segment PS ≅ segm
natima [27]

Answer:

Because the diagonals of a rectangle are congruent, the statement "segment SQ ≅ segment PR" is true.

5 0
2 years ago
Solve the system of equations.<br> <img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7By%3D220x-160%7D%20%5Catop%20%7By%3
Leni [432]

\begin{cases} y=220x-160\\ y=20x^2-400 \end{cases}\qquad \qquad \stackrel{\textit{substituting on the 2nd equation}}{\stackrel{\textit{\Large y}}{220x-160}~~ = ~~20x^2-400} \\\\\\ -160=20x^2-220x-400\implies 0=20x^2-220x-240 \\\\\\ 0=20(x^2-11x-12)\implies 0=x^2-11x-12 \\\\\\ 0=(x-12)(x+1)\implies x= \begin{cases} 12\\ -1 \end{cases}

\dotfill\\\\ x=12\qquad \qquad \stackrel{\textit{substituting on the 1st equation}}{y=220(12)-160}\implies y=2480 \\\\\\ x=-1\qquad \qquad \stackrel{\textit{substituting on the 1st equation}}{y=220(-1)-160}\implies y=-380 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (12~~,~~2480)\qquad (-1~~,~~-380)~\hfill

4 0
2 years ago
Look at the graph shown below:
ElenaW [278]
So.. take a peek at the picture... let's get two points from it, hmm say 0,4 notice it touches the y-axis there, and say hmmm -4, 1, almost at the bottom of the line

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\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad &#10;\begin{array}{llll}&#10;\textit{plug in the values for }&#10;\begin{cases}&#10;y_1=4\\&#10;x_1=0\\&#10;m=\boxed{?}&#10;\end{cases}\\&#10;\textit{and solve for "y"}&#10;\end{array}\\&#10;\left. \qquad   \right. \uparrow\\&#10;\textit{point-slope form}

once you get the slope and solve for "y", that'd be the equation of the line.
7 0
2 years ago
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