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Lana71 [14]
3 years ago
6

Please answer properly.

Mathematics
1 answer:
vagabundo [1.1K]3 years ago
3 0

Answer:

1) b=6

2) m=4

3) p=12

4) x=-1\frac{2}{6}

Step-by-step explanation:

b+6=12

subtract 6 from both sides

b=6

---------->>>>

9m=32+4

combine like terms

9m=36

divide both sides by 9

m=4

---------->>>>

21=p+9

Subtract 9 from both sides

12 = p

---------->>>>

(10+5)+3x=0

Add within the parenthesis

5+3x=0

subtract 5 from both sides

3x=-5

divide both sides by 3

x=-1\frac{2}{6}

Hope this is helpful to you.

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FACTORED POLYNOMIAL ONLY
Leto [7]

Answer:5x^{3}(x^{3}+2x+3)

3 0
3 years ago
Triangle abc is shown on the graph what are the coordinates of the image of point B after the triangle is rotated 270 about the
Lemur [1.5K]

Answer: (4,2)

Step-by-step explanation:

7 0
3 years ago
The coordinates below represent two linear equations.
allsm [11]
The answer is B. exactly 1.

As the given above the two line both have a point of (0,4).
The other two given point are different but that's okay because they're two different line with different slope just the same intercept.
5 0
3 years ago
Solve for x in the triangle. Round your answer to the nearest tenth.
kramer

Answer:

<h2>x = 10.3</h2>

Step-by-step explanation:

Use sine.

sine=\dfrac{opposite}{hypotenuse}

We have:

opposite=x\\hypotenuse=16

\sin40^o\approx0.6428        <em>look at the table in the picture</em>

Substitute:

\dfrac{x}{16}=0.6428         <em>multiply both sides by 16</em>

x=10.2848\to x\approx10.3

5 0
3 years ago
If (ax+2)(bx+7)=15x2+cx+14 for all values of x, and a+b=8, what are the 2 possible values fo c
dolphi86 [110]

Given:

(ax+2)(bx+7)=15x^2+cx+14

And

a+b=8

Required:

To find the two possible values of c.

Explanation:

Consider

\begin{gathered} (ax+2)(bx+7)=15x^2+cx+14 \\ abx^2+7ax+2bx+14=15x^2+cx+14 \end{gathered}

So

\begin{gathered} ab=15-----(1) \\ 7a+2b=c \end{gathered}

And also given

a+b=8---(2)

Now from (1) and (2), we get

\begin{gathered} a+\frac{15}{a}=8 \\  \\ a^2+15=8a \\  \\ a^2-8a+15=0 \end{gathered}a=3,5

Now put a in (1) we get

\begin{gathered} (3)b=15 \\ b=\frac{15}{3} \\ b=5 \\ OR \\ b=\frac{15}{5} \\ b=3 \end{gathered}

We can interpret that either of a or b are equal to 3 or 5.

When a=3 and b=5, we have

\begin{gathered} c=7(3)+2(5) \\ =21+10 \\ =31 \end{gathered}

When a=5 and b=3, we have

\begin{gathered} c=7(5)+2(3) \\ =35+6 \\ =41 \end{gathered}

Final Answer:

The option D is correct.

31 and 41

8 0
1 year ago
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