All categories

3 years ago

What is the factored form of 8x^2 + 12x

Mathematics

2 answers:

Crazy boy [7]3 years ago

6 0

The answer is the following:

4x(2x+3)

hope it helped :)

Vaselesa [24]3 years ago

5 0

Answer: The required factored form of the given expression is $4x(2x+3).$

Step-by-step explanation: We are given to find the factored form of the following quadratic expression :

$E=8x^2+12x~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Since the given quadratic expression does not contain the constant term, so first we should take common the x-term and then factorize.

From equation (i), we have

$E\\\\=8x^2+12x\\\\=4x(2x+3).$

Thus, the required factored form of the given expression is $4x(2x+3).$

You might be interested in

Help please!!!!!!!!!!!!!!!!!!! very important

qaws [65]

Answer:

i got B for my answer

4 0

3 years ago

A. Write 64√2 in the form 2^n

xxMikexx [17]

We get the expression 64√2 in the form 2ⁿ as 2⁶°⁵.

An expression is a way in which numbers are written.

It can be variables, constants or a combination of both.

We are given an expression:

64 √2

We know that:

64 = 2 × 2 × 2 × 2 × 2 × 2

64 = 2⁶

Also, we can write √2 as

√2 = 2⁰°⁵

We get that:

2⁶ × 2⁰°⁵

simplifying the expression, we get that:

= 2⁶⁺⁰°⁵

= 2⁶°⁵

Therefore, we get the expression 64√2 in the form 2ⁿ as 2⁶°⁵.

Learn more about expression here:

brainly.com/question/723406

#SPJ9

7 0

1 year ago

Domain of y=3secx/4 is

sladkih [1.3K]

Umm I didn't get a domain.... No solutions were found...

3 0

3 years ago

If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

4 years ago

A data set includes the following grades: 98, 94, 93, 82, 61, 80. Suppose that 61 is removed from the data. How is the median af

puteri [66]

Answer:

<u>If we remove 61 from the data set, the median changes from 87.5 to 93.</u>

Step-by-step explanation:

1. Let's calculate the median of the original data set:

Median = (3rd term + 4th term)/2 because the number of terms are even and our median mark is the average of the two middle marks, in this case, 82 and 93.

Median = (82 + 93)/2

Median = 87.5

2. Let's calculate the median of the data set removing 61:

Median = 3rd term because our median mark is the middle mark, in this case, 93. It is the middle mark because there are 2 scores before it (80 and 82) and 2 scores (94 and 98) after it.

Median = 93

7 0

3 years ago

Read 2 more answers