All categories

3 years ago

Please help asap. 45 points

Mathematics

2 answers:

Levart [38]3 years ago

5 0

The answer is B. hope that help

ch4aika [34]3 years ago

5 0

2(10x^2) + 22x -12

2(10x^2) + 2(11x) - 12

2(10x^2) + 2(11x) + 2 (-6)

2(10x^2 + 11x) + 2 (-6)

2 (10x^2 + 11x - 6)

2 (10x^2 - 4x + 15x - 6)

2 (2x (5x - 2) + 3 (5x - 2))

2 ((5x - 2) (2x + 3))

Finally, your answer is 2 (5x - 2) (2x + 3)

You might be interested in

24) Evaluate the expression with its given values. x4+3y;x=12,y=2

Minchanka [31]

$\large\text{Hey there!}$

$\mathsf{\dfrac{x}{4} + 3y}\\\\\mathsf{= \dfrac{12}{4} + 3(2)}\\\\\mathsf{= 3 + 3(2)}\\\\\mathsf{= 3 + 6}\\\\\mathsf{= 9}\\\\\large\text{Therefore, your answer: \huge\boxed{\mathsf{9}}}\huge\checkmark$

$\large\text{Good luck on your assignment \& enjoy your day!}$

~ $\frak{Amphitrite1040:)}$

8 0

3 years ago

Read 2 more answers

Chen has 4 pizzas to share with his friends. If each person will eat 2/3 of a pizza,how many people can share the pizzas?

sweet-ann [11.9K]

Answer:6 people

Step-by-step explanation: divide 4 by 2/3

4 0

4 years ago

Really easy points!!!!!! Please solve for all three answers.

katen-ka-za [31]

Answer:

1/64, 1/256, 1/1024

Step-by-step explanation:

To get from 16 to 4 we multiply by 1/4

To get from 4 to 1 we multiply by 1/4

Each time we multiply by 1/4

The next term would be 1/16 * 1/4

1/16 * 1/4 = 1/64

The take that term 1/64 and multiply by 1/4

1/64*1/4 = 1/256

Finally take 1/256 and multiply by 1/4

1/256*1/4 =1/1024

8 0

3 years ago

An equation that is true for every value of the variables is a(n)

Roman55 [17]

Identity, For example, x + x = 2x is true for every value of x.

5 0

3 years ago

Read 2 more answers

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago