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2 years ago

Please someone help me with this question

Mathematics

1 answer:

bagirrra123 [75]2 years ago

5 0

ANSWER_____9.2M_________

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Solve: 2 In 3 = In(x-4) x= 9 x = 10 x = 13 DONE

Airida [17]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

Name the identical triangles and measurement of side BA

aalyn [17]

Step-by-step explanation:

BE/BC=BD/BA

tHE TWO SIMILAR TRIANGLES ARE BED AND BC. tHE PROPORTION ABOVE REPRESENTS USE OF THE TWO SIMILAR TRIANGLES.

4/(4 + 10) = 5/(5+X) Cross Multiply

4(5+x) = 5(4+10)m

20 +4x=20 +50 subtract 20 from both sides

4x = 50 divide by 4

x =12.5

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3 years ago

Given the functions, f(x) =x2+x-5 and g(x)=4x2-2x+1 perform the indicated operation when applicable state the domain restriction

sashaice [31]

Your answer would be: -3x2 + 3x - 6.

I looked online to find this answer. All credits to their owners. Answer found on brainly.com/question/2825454.

This is not my answer. I'm just here to help.

4 0

3 years ago

What are the possible rational roots of the polynomial equation? 0=2x7+3x5−9x2+6

RoseWind [281]

Answer: $\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}$

Step-by-step explanation:

We can use the Rational Root Test.

Given a polynomial in the form:

$a_nx^n +a_{n- 1}x^{n - 1} + … + a_1x^1 + a_0 = 0$

Where:

- The coefficients are integers.

- $a_n$ is the leading coeffcient ( $a_n\neq 0$ )

- $a_0$ is the constant term $a_0\neq 0$

Every rational root of the polynomial is in the form:

$\frac{p}{q}=\frac{\pm(factors\ of\ a_0)}{\pm(factors\ of\ a_n)}$

For the case of the given polynomial:

$2x^7+3x^5-9x^2+6=0$

We can observe that:

- Its constant term is 6, with factors 1, 2 and 3.

- Its leading coefficient is 2, with factors 1 and 2.

Then, by Rational Roots Test we get the possible rational roots of this polynomial:

$\frac{p}{q}=\frac{\pm(1,2,3,6)}{\pm(1,2)}=\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}$

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3 years ago

Find the first five terms of the sequence given the following recursive formula:

forsale [732]

Answer:

The first five terms are;

-3,-7,11,-29,69

Step-by-step explanation:

The recursive definition of the sequence is

$a_1=-3,$ , $a_2=-7$ and $a_n=a_{n-2}-2a_{n-1}$ .

When n=3, we obtain;

$a_3=a_{3-2}-2a_{3-1}$ .

$\implies a_3=a_{1}-2a_{2}$ .

$\implies a_3=-3-2(-7)$ .

$\implies a_3=11$ .

When n=4

$\implies a_4=a_{2}-2a_{3}$ .

$\implies a_4=-7-2(11)=-29$ .

When n=5

$\implies a_5=a_{3}-2a_{4}$ .

$\implies a_4=11-2(-29)=69$ .

Therefore the first five terms are;

-3,-7,11,-29,69

5 0

3 years ago