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3 years ago

The product of the 2nd and 3rd term of a gp is 972.if the first term is 3/4.find the common ratio.

Mathematics

1 answer:

lana [24]3 years ago

4 0

Answer:

Common ratio = 12.

Step-by-step explanation:

2nd term = a1r and 3rd = a1r^2

So (a1)^2 r^3 = 972

a1 = 3/4, so

(3/4)^2 r^3 = 972

r^3 = 972 / 9/16

r^3 = 972*16/9 = 1728

r = 12.

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Determine the least possible degree of the polynomial function shown

OleMash [197]

Answer:

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Step-by-step explanation:

To find the degree of the polynomial, add up the exponents of each term and select the highest sum.So the least number of roots = 5 and this will also be the least possible degree of the polynomial

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3 years ago

Find the solution to the following system using substitution or elimination: y = 3x + 2 y = -2x - 8 O A. (-5,-) OB. (-2,-4) O C.

mr Goodwill [35]

Answer:

B. (-2,-4)

Explanation

Given equations:

y = 3x + 2

y = -2x - 8

Solving both equations will yield the values of x and y;

Solution:

y = 3x + 2 ----- (i)

y = -2x - 8 ------ (ii)

Using substitution method, input equation i, into ii

3x + 2 = -2x - 8

Collect like terms and solve;

3x + 2x = -8 -2

5x = -10

x = -2

Then put x = -2 into i, to find y

y = (-2 x 3) + 2

y = -6 + 2 = -4

So, the solution of the equation is B. (-2,-4)

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3 years ago

Solving special systems.

goldfiish [28.3K]

Answer:

Step-by-step explanation:

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3 years ago

I need help on ratio of x-coordinates and ratio of y-coordinates

ad-work [718]

Answer:

Ratio of x-coordinates:

$\frac{-1}{-2}=\frac{1}{2}$

$\frac{2}{4}=\frac{1}{2}$

$\frac{-2}{-4}=\frac{1}{2}$

Ration of y-coordinates:

$\frac{2.5}{5}=\frac{1}{2}$

$\frac{1}{2}=\frac{1}{2}$

$\frac{-2}{-4}=\frac{1}{2}$

Step-by-step explanation:

The table is asking for the ratio of x-coordinates for each point (A, B, C and D) for both the image and pre-image. The ratio is the image 'x' or 'y' value ÷ the pre-image 'x' or 'y' value. Each ratio should be expressed in simplest form and should show the same pattern of dilation, or same scale factor. In this case, the second figure is 1/2 the size of the original figure.

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3 years ago

Please help ASAP!

Kruka [31]

First, recall that

$\sin(-\theta)=-\sin\theta$

So if $\sin(-\theta)=\dfrac15$ , then $\sin\theta=-\dfrac15$ .

Second,

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

We know that $\tan\theta>0$ and $\sin\theta$ , which means we should also have $\cos\theta$ .

Third,

$\sin^2\theta+\cos^2\theta=1\implies\cos\theta=\pm\sqrt{1-\sin^2\theta}$

but as we've already shown, we need to have $\cos\theta$ , so we pick the negative root.

Finally,

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}\iff\dfrac6{\sqrt{12}}=\dfrac{-\frac15}{\cos\theta}\implies\cos\theta=-\dfrac1{5\sqrt3}$

Unfortunately, none of the given answers match, so perhaps I've misunderstood one of the given conditions... In any case, this answer should tell you everything you need to know to find the right solution from the given options.

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3 years ago