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Blababa [14]
3 years ago
15

Find the solutions to (x+4)^2=36.

Mathematics
2 answers:
KATRIN_1 [288]3 years ago
8 0
The answers are -10 and 2
JulsSmile [24]3 years ago
5 0

Step-by-step explanation:

So here are the two solutions...

A.-10

B.6

Hope it helps you

You might be interested in
How many cubes of side 3 cm can be cut from a wooden solid cuboid with dimensions 12 cm x 12 cm x 9 cm?
Orlov [11]

Answer:

48 cubes can be cut from the wooden cuboid

Step-by-step explanation:

Total calculate this, we will first of all find the volume of the cuboid, and the volume of the cubes to be cut, then divide the volumes to see how many cubes can be cut from the cuboid.

Volume of cuboid = 12 × 12 × 9 = 1296 cm

volumes of each cube to be cut = 3 × 3 × 3 = 27 cm

Next, we will divide the volume of cuboid by the volume of the cubes:

Number of cubes = Volume of cuboid ÷ volume of cubes

Number of cubes = 1296 ÷ 27 = 48 cubes

Therefore, 48 cubes of sides 3cm can be cut from the wooden cuboid

5 0
3 years ago
5(3r+11)=85
raketka [301]

Answer:

r = 2

Step-by-step explanation:

Actually, there are 2 ways to solve this equation.

The first one:

<em>5</em><em>(</em><em> </em><em>3</em><em>r</em><em> </em><em>+</em><em> </em><em>1</em><em>1</em><em>)</em><em> </em><em>=</em><em> </em><em>8</em><em>5</em>

<em>so </em><em>here</em><em>, </em><em> </em><em>you </em><em>have </em><em>to </em><em>multiply </em><em>5</em><em> </em><em>by </em><em>(</em><em> </em><em>3</em><em>r</em><em> </em><em>+</em><em> </em><em>1</em><em>1</em><em>)</em><em> </em><em>and </em><em>maintain </em><em>the </em><em>8</em><em>5</em>

<em>1</em><em>5</em><em>r</em><em> </em><em>+</em><em> </em><em>5</em><em>5</em><em> </em><em>=</em><em> </em><em>8</em><em>5</em><em>.</em><em>.</em><em>.</em><em> </em><em> </em><em> </em><em>now </em><em>put </em><em>the </em><em>like </em><em>terms </em><em>together</em><em>.</em><em> </em><em>In </em><em>other </em><em>words</em><em>, </em><em>just </em><em>transpose </em><em>5</em><em>5</em><em> </em><em>to </em><em>the </em><em>other </em><em>side. </em>

<em>1</em><em>5</em><em>r</em><em> </em><em>=</em><em> </em><em>8</em><em>5</em><em> </em><em>-</em><em>5</em><em>5</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em> </em><em>remember </em><em>that </em><em>the </em><em>sign </em><em>of </em><em>5</em><em>5</em><em> </em><em>will </em><em>change </em><em>from </em><em>positive </em><em>to </em><em>negative </em>

<em>1</em><em>5</em><em>r</em><em> </em><em>=</em><em> </em><em>3</em><em>0</em><em>.</em><em>.</em><em>.</em><em>.</em><em> </em><em>now </em><em>divide </em><em>both </em><em>sides </em><em>by </em><em> </em><em>1</em><em>5</em>

<em>1</em><em>5</em><em>r</em><em>/</em><em>1</em><em>5</em><em> </em><em>=</em><em> </em><em>3</em><em>0</em><em>/</em><em>1</em><em>5</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em><u>r </u></em><em><u>=</u></em><em><u> </u></em><em><u>2</u></em>

The second method :

<em>5</em><em> </em><em>(</em><em> </em><em>3</em><em>r</em><em> </em><em>+</em><em> </em><em>1</em><em>1</em><em>)</em><em> </em><em>=</em><em> </em><em>8</em><em>5</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em> </em><em>divide </em><em>both </em><em>sides </em><em>by </em><em>5</em>

<em>5</em><em> </em><em>(</em><em>3</em><em>r</em><em> </em><em>+</em><em> </em><em>1</em><em>1</em><em>)</em><em> </em><em>/</em><em> </em><em>5</em><em> </em><em>=</em><em> </em><em>8</em><em>5</em><em>/</em><em>5</em>

<em>3</em><em>r</em><em> </em><em>+</em><em> </em><em>1</em><em>1</em><em> </em><em>=</em><em> </em><em>1</em><em>7</em>

<em>3</em><em> </em><em>r </em><em>=</em><em> </em><em>1</em><em>7</em><em> </em><em>-</em><em> </em><em>1</em><em>1</em>

<em>3</em><em>r</em><em> </em><em>=</em><em> </em><em>6</em><em> </em>

<em>r </em><em>=</em><em> </em><em>2</em>

<em>I </em><em>hope </em><em>this </em><em>helps. </em><em>If </em><em>you </em><em>don't </em><em>quite </em><em>get </em><em>it, </em><em> </em><em>please </em><em>go </em><em>through </em><em>until </em><em>you </em><em>get </em><em>it. </em><em> </em><em>Have </em><em>a </em><em>nice </em><em>day </em>:-)

5 0
3 years ago
Read 2 more answers
11/12 divided by 1/3<br>in fraction form<br><br>​
SashulF [63]

Answer:

11/4 or 2 3/4

Step-by-step explanation:

Step 1:

11/12 ÷ 1/3         Equation

Step 2:

11/12 × 3/1    Flip the fraction

Step 3:

11/4 × 1/1      Simplify

Step 4:

11/4      Multiply

Answer:

11/4 or 2 3/4

Hope This Helps :)

7 0
3 years ago
Read 2 more answers
Solve and show steps. will award brainliest.
a_sh-v [17]

For the first question please find the attached diagram.

As per the diagram, P is the upstream point and Q is the downstream point. The distance between P and Q is 22.5 miles.

Let the speed of the boat in the still waters of the lake be represented by S.

Then, when the boat travels upstream, the net speed of the boat will be (S-6) miles per hour because the river flows downstream and thus the speed of the boat will have to be subtracted from the speed of the river.

Now, we know that the relationship between the net speed, distance and time of travel is give as:

Distance = Net Speed x Time of travel

For the upstream ride of the board we know that Distance is 22.5 miles and Net Speed is (S-6). Therefore, the above equation will become:

22.5=(S-6)\times T_{1} where T_{1} represents the time taken to travel upstream.

We can rearrange the above equation to be:

T_{1}=\frac{22.5 }{S-6}......................(Equation 1)

By similar arguments we know that the downstream speed of the boat is S+6 and the distance travelled is the same and so the time taken to travel downstream (represented by T_{2}) will be:

T_{2}=\frac{22.5}{S+6}................(Equation 2)

Now, we know that the total time of travel should be 9 hours.

This means that: T_{1}+T_{2}=9............(Equation 3)

Plugging in the values of T_{1} and T_{2} from (Equation 1) and (Equation 2) into (Equation 3), we get:

\frac{22.5 }{S-6} +\frac{22.5 }{S+6}=9

Simplifying the above we will get a quadratic equation:

9S^2-45S-54=0

The roots of this quadratic equation are:

S=-1 and S=6

Since, speed cannot be negative, S=-1 is out of consideration.

The speed of the boat in the lake is thus S=6 miles per hour.

But we have a problem with S=6 too. The problem is that if S=6, then the boat will not be able to move upstream.

Let us solve problem 2

We are given that: \frac{x-2}{x+3}+\frac{10x}{x^2-9}

We can rewrite it as:

\frac{x-2}{x+3}+\frac{10x}{(x-3)(x+3)}

\frac{(x-3)(x-2)+10x}{(x-3)(x+3)} =\frac{x^2-5x+6+10x}{(x-3)(x+3)}

Now, the numerator can be simplified as:

\frac{x^2+10x+6}{(x-3)(x+3)} =\frac{(x+3)(x+2)}{(x-3)(x+3)} =\frac{x+2}{x-3}

Thus, our final simplified answer is:

\frac{x+2}{x-3}

The restriction on the variable x is that it cannot be equal to either +3 or -3 as that would make the denominator of the original question equal to zero.

Thus, the restriction is x\neq \pm 3

8 0
4 years ago
Find the solutions of :<br> -3X + 2 2 14<br> Give 5 solutions
pshichka [43]

Answer:

-3(x - 738)

Step-by-step explanation:

-3x+2214

-3x+3 . 738

-3(x-738)

5 0
3 years ago
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