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2 years ago

Solve by factoring 6x^2 +13x -28 =0

Mathematics

2 answers:

Sauron [17]2 years ago

8 0

Answer:

x=4/3 or x=-7/2

Step-by-step explanation:

(3x-4)(2x+7)

Alik [6]2 years ago

4 0

Answer:

x=-7/2, x=4/3

Step-by-step explanation:

$6x^2+13x-28=0$

Multiply 6 and -28 to get -168

Find 2 numbers that multiply to 168 but add to 13

They are 21 & -8

Rewrite the equation into:

$6x^2+21x-8x-28=0$

Factor

$3x(2x+7)-4(2x+7)=0$

(3x-4)(2x+7)=0

3x=4

x=4/3

2x=-7

x=-7/2

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3 years ago

Tyler buys 3 1/2 pounds of fresh beans. The beans are on sale for $0.98 per pound. Which expressions can Tyler use to calculate

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Step-by-step explanation:

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3 years ago

Which of these sets could represent the side lengths of a right triangle?

Jlenok [28]

Answer:

{6, 8, 10} is a set which represents the side length of a right triangle.

Step-by-step explanation:

In a right triangle:

$(Base)^{2} + (Perpendicular)^{2} = (Hypotenuse)^{2}$

Now, in the given triplets:

(a) {4, 8, 12}

Here, $(4)^{2} + (8)^{2} = 16 + 64 = 80\\\implies H = \sqrt{80} = 8.94$

So, third side of the triangle 8.94 ≠ 12

Hence, {4, 8, 12} is NOT a triplet.

(b) {6, 8, 10}

Here, $(6)^{2} + (8)^{2} = 36 + 64 = 100\\\implies H = \sqrt{100} = 10$

So, third side of the triangle 10

Hence, {6, 8, 10} is a triplet.

Here, $(6)^{2} + (8)^{2} = 36 + 64 = 100\\\implies H = \sqrt{100} = 10$

So, third side of the triangle 10 ≠ 15

Hence, {6, 8, 15} is NOT a triplet.

(d) {5, 7, 13}

Here, $(5)^{2} + (7)^{2} = 25 + 49 = 74\\\implies H = \sqrt{74} = 8.60$

So, third side of the triangle 8.60 ≠ 13

Hence, {5, 7, 13} is NOT a triplet.

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3 years ago

<img src="https://tex.z-dn.net/?f=%20-%203%20%2B%205i%20%20%5Cdiv%20%20-%203%20-%204i" id="TexFormula1" title=" - 3 + 5i \div

Likurg_2 [28]

Answer:

$\frac{-11}{25}+\frac{-27}{25}i$ given you are asked to simplify

$\frac{-3+5i}{-3-4i}$

Step-by-step explanation:

You have to multiply the numerator and denominator by the denominator's conjugate.

The conjugate of a+bi is a-bi.

When you multiply conjugates, you just have to multiply first and last.

(a+bi)(a-bi)

a^2-abi+abi-b^2i^2

a^2+0 -b^2(-1)

a^2+-b^2(-1)

a^2+b^2

See no need to use the whole foil method; the middle terms cancel.

So we are multiplying top and bottom of your fraction by (-3+4i):

$\frac{-3+5i}{-3-4i} \cdot \frac{-3+4i}{-3+4i}=\frac{(-3+5i)(-3-4i)}{(-3-4i)(-3+4i)}$

So you will have to use the complete foil method for the numerator. Let's do that:

(-3+5i)(-3+4i)

First: (-3)(-3)=9

Outer:: (-3)(4i)=-12i

Inner: (5i)(-3)=-15i

Last: (5i)(4i)=20i^2=20(-1)=-20

--------------------------------------------Combine like terms:

9-20-12i-15i

Simplify:

-11-27i

Now the bottom (-3-4i)(-3+4i):

F(OI)L (we are skipping OI)

First:-3(-3)=9

Last: -4i(4i)=-16i^2=-16(-1)=16

---------------------------------------------Combine like terms:

9+16=25

So our answer is $\frac{-11-27i}{25}{/tex] unless you want to seprate the fraction too:[tex]\frac{-11}{25}+\frac{-27}{25}i$

6 0

3 years ago