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3 years ago

HELP PLEASE FAST PLEASE HELP!!!!!!!! Solve for x: one over five (5x + 12) = 18

Mathematics

1 answer:

tiny-mole [99]3 years ago

7 0

X=6/5 change the addition sign to subtraction then subtract then divide by 5 and it is 6/5

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An artist wants to frame a square painting with an area of 400 square inches. She wants to know the length of the wood trim that

Maurinko [17]

Area=legnth times width

if legnth=x
and
area=400

400=x times width

wait
SQUARE painting
legnth=width
width=x

400=x times x
400=x^2
sqrt both sides
20=x
answer is 20 in

equation is
400=x times x

5 0

3 years ago

The sum of the areas of two circles is 8pie square meters. Find the length of a radius of each circle ifone of them is twice as

NeX [460]

Nice

area=pir^2
the areas are
pir^2+pi(2r)^2
they add to 8pi

8pi=pir^2+pi(2r)^2
undistribute pi
8pi=pi(r^2+(2r)^2)
divide both sides by pi
8=r^2+(2r)^2
expand
8=r^2+4r^2
add
8=5r^2
divide by 5
8/5=r^2
sqrt both sides
√(8/5)=r
double since it is double
2√(8/5) is radius of bigger aka (4√10)/5

radius=(4√10)/5

5 0

3 years ago

What is -3 = z - 8 (solve the equation, check the solution)

laila [671]

Answer:

z=5 :)

Step-by-step explanation:

5 0

3 years ago

Two buses leave towns 304 miles apart at the same time and travel toward each other. One bus travels 14 mih slower than the othe

8090 [49]

Answer:

The faster bus moves at 83mi/h and the slower one moves at 69mi/h.

Step-by-step explanation:

Let's define:

R₁ = rate of bus 1, this is the faster one.

R₂ = rate of bus 2, this is the slower one.

We know that one bus travels 14mi/h slower, then:

R₂ = R₁ - 14mi/h.

Now we know that:

Distance = Speed*Time.

If we add the distances that both busses travel in 2 hours, it should be equal to the initial distance between the buses, then:

R₁*2h + R₂*2h = 304 mi

Then we have the two equations:

R₂ = R₁ - 14mi/h

R₁*2h + R₂*2h = 304 mi

The first step is to replace the first equation in the second one, to get:

R₁*2h + (R₁ - 14mi/h)*2h = 304 mi

And now we can solve this for R₁.

R₁*2h + R₁*2h - 14mi/h*2h = 304 mi

R₁*4h - 28mi = 304mi

R₁*4h = 304mi + 28mi = 332mi

R₁ = 332mi/4h = 83mi/h

The faster bus moves at 83mi/h

And we know that the slower one moves at 14mi/h slower than this, then:

R₂ = R₁ - 14mi/h = 83mi/h - 14mi/h = 69 mi/h

7 0

3 years ago

Given that Cosecant (t) = negative StartFraction 13 Over 5 EndFraction for Pi less-than t less-than StartFraction 3 pi Over 2 En

Sergio [31]

Answer:

$\bold{cot(t) =\dfrac{12}{5}}$

Step-by-step explanation:

Given that:

$Cosec (t) = -\frac{13}5$

for $\pi < t < \frac{3 \pi}2$

That means, angle $t$ is in the 3rd quadrant.

To find:

Value of cot(t)

Solution:

First of all, let us recall what trigonometric ratios are positive and what trigonometric ratios are negative in 3rd quadrant.

In 3rd quadrant, tangent and cotangent are positive.

All other trigonometric ratios are negative.

Let us have a look at the following identity:

$cosec^2\theta -cot^2\theta =1$

here, $\theta =t$

So, $cosec^2t-cot^2t=1$

$\Rightarrow (-\dfrac{13}{5})^2-cot^2t=1\\\Rightarrow (\dfrac{169}{25})-cot^2t=1\\\Rightarrow \dfrac{169}{25}-1=cot^2t\\\Rightarrow \dfrac{169-25}{25}=cot^2t\\\Rightarrow \dfrac{144}{25}=cot^2t\\\Rightarrow cot(t)=\pm\sqrt{\dfrac{144}{25}}\\\Rightarrow cot(t)=\pm\dfrac{12}{5}$

But, angle $t$ is in 3rd quadrant, so value of

$\bold{cot(t) =\dfrac{12}{5}}$

4 0

3 years ago