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3 years ago

Consider this expression

Mathematics

2 answers:

Sloan [31]3 years ago

8 0

Answer:

$\displaystyle \large{(x + 9)(x - 8)}$

The value of a is 9 and b is -8

Step-by-step explanation:

$\displaystyle \large{(x + a)(x + b) = {x}^{2} + (a + b)x + ab}$

We need to find two numbers that add up or subtract off to 1.

Then we also use the same two numbers that multiply and get -72.

Let's see:-

9 and -8 seem like correct values.

Because 9+(-8) is 9-8 = 1

And 9(-8) is -72.

Therefore:

$\displaystyle \large{ {x}^{2} + x - 72 = (x + 9)(x - 8)}$

The value of a is 9 and the value of b is -8 according to form of (x+a)(x+b)

Bess [88]3 years ago

6 0

Answer:

(x + 9)(x - 8)

Step-by-step explanation:

x² + x - 72

Consider the factors of the constant term (- 72) which sum to give the coefficient of the x- term (+ 1)

The factors are + 9 and - 8 , since

9 × - 8 = - 72 and 9 - 8 = 1 , then

x² + x - 72 = (x + 9)(x - 8)

with a = 9 and b = - 8

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Based on its degree, what kind of polynomial is this h(x)= -6x^3+2x-5

Lelechka [254]

Answer:

cubic polynomial

Step-by-step explanation:

Given polynomial is $h(x)=-6x^{3}+2x-5$

A polynomial of degree 1 is a linear polynomial.

A polynomial of degree 2 is a quadratic polynomial.

A polynomial of degree 3 is a cubic polynomial.

In this case the exponent with the maximum value in the polynomial is 3.

Hence the degree of the polynomial h(x) is 3.

Hence the given polynomial is a cubic polynomial.

7 0

3 years ago

Find the area of the figure. Use 3.14 as π.

Karolina [17]

Answer: D. 94.25 in²

Step-by-step explanation:

To find the total area, we will break the shape up into two different parts.

[] The rounded part is 39.25 in². Let us assume the rounded part is exactly half of a circle.

Area of a circle:

A = πr²

Use 3.14 for pi:

A = (3.14)r²

Find the radius:

d / 2 = r, 10 / 2 = 5 in

Subsittue:

A = (3.14)(5)²

A = 78.5 in²

Divide by 2 since it is only half:

78.5 in² / 2 = 39.25 in²

[] The triangle is 55 in².

Area of a triangle:

A = b*h/2

A = 11 * 10 / 2

A = 110 / 2

A = 55 in²

[] Total area. We will add the two parts together.

55 in² + 39.25 in² = 94.25 in²

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2 years ago

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-4(-2z + 2) = 6z + 8

Savatey [412]

8z-8=6z+8
8x-6z=8+8
2z=16
Z=8

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2 years ago

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Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

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3 years ago

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Find the product. (do not use spaces in your answer.) d ^j · d ^k

bezimeni [28]

$d^j\cdot d^k=d^{j+k}$

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3 years ago

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