Ask question

Ask question

All categories

2 years ago

12

Determine how many solutions the following equation has using quadratic formula x^2+5x=-8

1 answer:

Katen [24]2 years ago

3 0

Answer:

I will be using differentiation method

Step-by-step explanation:

x^2+5x =-8

2×x ×x^2-1 +5x = -8

2x +5x = -8

7x = -8

divide both sides by the co-efficient of x

7x÷7=-8÷7

x =-1.5

You might be interested in

will got 17 out of 25 questions correct on his driving test what percent correct did he get on the test?

Natalija [7]

Answer:

68%

Step-by-step explanation:

$\frac{17}{25} = \frac{68}{100}$
$\frac{68}{100}$ = 68%

I hope this helps!

5 0

3 years ago

Read 2 more answers

Marta and Kim are sisters. Five years ago Kim's age was twice as great as martas age. If Marta is no m years old, what expressio

Hitman42 [59]

Answer:

y = 2m-5

Kim's age now is y = 2m-5

Step-by-step explanation:

Let m represent martas age now and y represent Kim's age now

Given;

Five years ago Kim's age was twice as great as martas age.

(y-5) = 2(m-5)

Solving for y;

y-5 = 2m-10

y = 2m-10+5

y = 2m-5

Kim's age now is y = 2m-5

5 0

2 years ago

Pls help, will give brainliest

Alexeev081 [22]

Answer:

see image

Step-by-step explanation:

Hope this is clear enough, if not feel free to ask me any questions.

3 0

2 years ago

Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!

LUCKY_DIMON [66]

Answer:

$=8\sqrt{15}b^{\frac{7}{2}}$

Step-by-step explanation:

$\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}$

$=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}$

$=\sqrt{40}b^2\sqrt{24b^3}$

$\sqrt{24b^3}$

$=\sqrt{24}\sqrt{b^3}$

$=\sqrt{24}b^{\frac{3}{2}}$

$=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$\sqrt{2^3}$

$=2^{3\cdot \frac{1}{2}$

$=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}$

$2^{\frac{3}{2}+\frac{3}{2}}$

$=2^3$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}$

$b^{\frac{3}{2}+2}$

$=b^{\frac{7}{2}}$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}$

$=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}$

$=8\sqrt{15}b^{\frac{7}{2}}$

4 0

3 years ago

How can we categorize colloids?

solniwko [45]

Step-by-step explanation:

a stable system of two phases one of which is dispersed in other form of particles

7 0

2 years ago

Read 2 more answers