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3 years ago

Which two values of x are roots of the polynomial below

Mathematics

1 answer:

katrin [286]3 years ago

7 0

There are no roots for this quadratic

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The sum of 9 and a number is twice a number plus 14. How would I write this as an equation?

valina [46]

1+2+3+4+5+6+7+8+9×2+14

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3 years ago

Determine the ratio in which the point (–6, m) divides the join of A(–3, –1) and B(–8, 9). Also, find the value of m.

PSYCHO15rus [73]

Answer:

Ratio = 3 : 2 and value of m = 5.

Step-by-step explanation:

We are given the end points ( -3,-1 ) and ( -8,9 ) of a line and a point P = ( -6,m ) divides this line in a particular ratio.

Let us assume that it cuts the line in k : 1 ratio.

Then, the co-ordinates of P = $( \frac{-8k-3}{k+1},\frac{9k-1}{k+1} )$ .

But, $\frac{-8k-3}{k+1}$ = -6

i.e. -8k-3 = -6k-6

i.e. -2k = -3

i.e. $k = \frac{3}{2}$

So, the ratio is k : 1 i.e $\frac{3}{2} : 1$ i.e. 3 : 2.

Hence, the ratio in which P divides the line is 3 : 2.

Also, $\frac{9k-1}{k+1}$ = m where $k = \frac{3}{2}$

i.e. m = $\frac{\frac{9 \times 3}{2}-1}{\frac{3}{2}-1}$

i.e. m = $\frac{27-2}{3+2}$

i.e. m = $\frac{25}{5}$

i.e. m = 5.

Hence, the value of m is 5.

4 0

3 years ago

Read 2 more answers

What is the hypothesis in this conditional statement?

nexus9112 [7]

The answer is
A.
Hope This Helps

8 0

3 years ago

16) Please help with question. WILL MARK BRAINLIEST + 10 POINTS.

Katyanochek1 [597]

We will use the sine and cosine of the sum of two angles, the sine and consine of $\frac{\pi}{2}$ , and the relation of the tangent with the sine and cosine:

$\sin (\alpha+\beta)=\sin \alpha\cdot\cos\beta + \cos\alpha\cdot\sin\beta

\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$

$\sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0$

$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$

If you use those identities, for $\alpha=x,\ \beta=\dfrac{\pi}{2}$ , you get:

$\sin\left(x+\dfrac{\pi}{2}\right) = \sin x\cdot\cos\dfrac{\pi}{2} + \cos x\cdot\sin\dfrac{\pi}{2} = \sin x\cdot0 + \cos x \cdot 1 = \cos x$

$\cos\left(x+\dfrac{\pi}{2}\right) = \cos x \cdot \cos\dfrac{\pi}{2} - \sin x\cdot\sin\dfrac{\pi}{2} = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$

Hence:

$\tan \left(x+\dfrac{\pi}{2}\right) = \dfrac{\sin\left(x+\dfrac{\pi}{2}\right)}{\cos\left(x+\dfrac{\pi}{2}\right)} = \dfrac{\cos x}{-\sin x} = -\cot x$

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3 years ago

Divide. Write your answer using the smallest numbers possible.

Ilya [14]

Answer:

30660

Step-by-step explanation:

8 0

2 years ago