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kenny6666 [7]
3 years ago
13

5 less than twice a number is 27. what’s the number

Mathematics
2 answers:
Vaselesa [24]3 years ago
7 0

Answer:

i believe that would be 11

Step-by-step explanation:

Nesterboy [21]3 years ago
3 0
Answer:
32

Explanation:
Add 5 to 27, it is 32. And to
Check that you divide 32 by 2, and you get 16 which is a whole number so that is how the answer is 32.
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Find the constant of proportionality if y is directly related to x when y = 12 and x = 6
Sergeeva-Olga [200]

Answer:

2

Step-by-step explanation:

Please see attached picture for full solution.

3 0
4 years ago
Please help: solve for x
klemol [59]

Answer:

x=11

Step-by-step explanation:

This is because 6+10=16 so your line with x must equal 16. You subtract 5 from 16 to get 11. This is your answer.

8 0
3 years ago
Q5. A car is bought for £5000 and sold for £7500.<br> What is the percentage profit?
Nitella [24]

Answer:

50% profit

Step-by-step explanation:

Cost price = £ 5000

Selling price = £ 7500

Profit  = Selling price - cost price

        = 7500 - 5000

       = £ 2500

Profit percentage = \frac{2500}{5000}*100

                             = 50 %

8 0
3 years ago
Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive
VikaD [51]
Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}&#10;\\\\\\&#10;\textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

recall that 

\bf sin(\theta)=\cfrac{opposite}{hypotenuse}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{adjacent}{hypotenuse}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{opposite}{adjacent}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{adjacent}{opposite}&#10;\\\\\\&#10;% cosecant&#10;csc(\theta)=\cfrac{hypotenuse}{opposite}&#10;\qquad \qquad &#10;% secant&#10;sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

\bf sin(\theta)=\cfrac{-1}{6}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{\sqrt{35}}{6}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{-1}{\sqrt{35}}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{\sqrt{35}}{1}&#10;\\\\\\&#10;sec(\theta)=\cfrac{6}{\sqrt{35}}

now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}&#10;\\\\\\&#10;sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
3 0
4 years ago
How can I tell, when doing synthetic division, if the binomial is<br> a FACTOR of the dividend?
Alecsey [184]

If you get 0 as the last value in the bottom row, then the binomial is a factor of the dividend.

Let's say the binomial is of the form (x-k) and it multiplies with some other polynomial q(x) to get p(x), so,

p(x) = (x-k)*q(x)

If you plug in x = k, then,

p(k) = (k-k)*q(k)

p(k) = 0

The input x = k leads to the output y = 0. Therefore, if (x-k) is a factor of p(x), then x = k is a root of p(x).

It turns out that the last value in the bottom row of a synthetic division table is the remainder after long division. By the remainder theorem, p(k) = r where r is the remainder after dividing p(x) by (x-k). If r = 0, then (x-k) is a factor, p(k) = 0, and x = k is a root.

5 0
3 years ago
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