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3 years ago

What is the answer for this equation? -2(3x+10)+14x=28

Mathematics

2 answers:

azamat3 years ago

8 0

Answer:

x= 6

Step-by-step explanation:

Distribute -2 into the parenthesis.

-6x+(-20)+14x= 28

The negative sign cancels out the addition sign,

-6x-20+14x= 28

Add the variables together.

(-6x+14)

8x-20= 28

Add 20 to both sides.

8x-20= 28

+20 +20

Divide to both sides.

2 2

x= 6

Alik [6]3 years ago

7 0

x=6 hi hope this helps.

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Simplify by expressing with radicals.

slamgirl [31]

Answer:

$6xy^3$

Step-by-step explanation:

$6\sqrt{x^2y^6}= \\\\6(x^{2/2}y^{6/2})= \\\\6xy^3$

Hope this helps!

6 0

3 years ago

Read 2 more answers

Find a polynomial of the form

Sloan [31]

Answer:

$f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3$

Step-by-step explanation:

We are given that

$f(x)=ax^3+bx^2+cx+d$

f(0)=-3,f(1)=2,f(3)=5 and f(4)=0

We have to find the polynomial

Substitute the value x=0 then ,we get

f(0)=d=-3

Substitute x=1 then we get

$a+b+c-3=2$

$a+b+c=2+3=5$

$a+b+c=5$ (equation I)

Substitute x=3 then we get

$a(3)^3+b(3)^2+c(3)-3=5$

$27a+9b+3c=5+3=8$

$27a+9b+3c=8$ (Equation II)

Substitute x=4 then we get

$a(4)^3+b(4)^2+c(4)-3=0$

$64a+16b+4c=3$ (Equation III)

Equation I multiply by 3 then subtract from equation II

$24a+6b=-7$ (Equation IV)

Equation II multiply by 4 and equation III multiply by 3 and subtract equation II from III

$84a+12b=-23$ (Equation V)

Equation IV multiply by 2 and then subtract from equation V

$36a=-9$

$a=-\frac{9}{36}$

$a=-\frac{1}{4}$

Substitute the value of a in equation IV then we get

$24(-\frac{1}{4})+6b=-7$

$-6+6b=-7$

$6b=-7+6=-1$

$b=-\frac{1}{6}$

Substitute the value of b in equation I then we get

$-\frac{1}{4}-\frac{1}{6}+ c=5$

$-\frac{5}{12}+c=5$

$c=5+\frac{5}{12}=\frac{65}{12}$

Substitute the values then we get

$f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3$

5 0

4 years ago

Marianna [84]

The expression into a single logarithm is $log[(x)^{10}][(2)^{30}]$

Step-by-step explanation:

Let us revise some logarithmic rules

$log(a)^{n}=nlog(a)$
$log(ab)=log(a)+log(b)$
$nlog(a)+mlog(b)=log[(a)^{n}][(b)^{m}]$

∵ 10 log(x) + 5 log(64)

- At first re-write 10 log(x)

∴ 10 log(x) = $log(x)^{10}$

- Then re-write 5 log(64)

∴ 5 log(64) = $log(64)^{5}$

∴ 10 log(x) + 5 log(64) = $log(x)^{10}$ + $log(64)^{5}$

- Use the 3rd rule above to make it single logarithm

∵ $log(x)^{10}$ + $log(64)^{5}$ = $log[(x)^{10}][(64)^{5}]$

∴ 10 log(x) + 5 log(64) = $log[(x)^{10}][(64)^{5}]$

∵ 64 = 2 × 2 × 2 × 2 × 2 × 2

∴ We can write 64 as $2^{6}$

∴ $(64)^{5}=(2^{6})^{5}$

- Multiply the two powers of 2

∴ $(64)^{5}=(2)^{30}$

∴ 10 log(x) + 5 log(64) = $log[(x)^{10}][(2)^{30}]$

The expression into a single logarithm is $log[(x)^{10}][(2)^{30}]$

Learn more:

You can learn more about the logarithmic functions in brainly.com/question/11921476

#LearnwithBrainly

6 0

3 years ago

Plz help its timedddddddddddddddddddd

stepladder [879]

Answer: (0,0),(0,9),(9,0)

Multiply each column by 3 to find 3T of Matrix.

6 0

3 years ago

Find the quotient of 2.9x10 and 4x10.-9

Natalka [10]

Answer:

2.9×10 = 29

4×10.-9 =31

8 0

3 years ago