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3 years ago

How do you solve (-3)(-1)

Mathematics

1 answer:

inysia [295]3 years ago

5 0

(-3)(-1)

- and - = + positive

(-3)(-1)

Answer: 3

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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

What is 4/7 x 3/4 and 1 2/3 x 1/5?

Elena-2011 [213]

Answer: 3/7 and 12/15

5 0

4 years ago

Read 2 more answers

The function f(x) = 1/2 x + 3/2 is used to complete this table.

Mrac [35]

Answer:

a) f(-1/2) = -2 is NOT TRUE.

b) f(0) =3/2 is TRUE.

c) f(1) = -1 is NOT TRUE.

d) f(2) = 1 is NOT TRUE.

e) f(4) = 7/2 is TRUE.

Step-by-step explanation:

Here, the given function is $f(x) = (\frac{1}{2}) x+\frac{3}{2}$

Now, checking for each values for the given function:

a) Putting x = (-1/2):

$f(\frac{-1}{2} ) = (\frac{1}{2})(\frac{-1}{2} ) +\frac{3}{2} = \frac{-1}{4} + (\frac{3}{2} )\\\implies f(x) = \frac{-1 + 6}{4} = (\frac{5}{4} )$

and (5/4) ≠ -2

Hence, f(-1/2) = -2 is NOT TRUE.

b)Putting x = 0 :

$f(0) = (\frac{1}{2})(0 ) +\frac{3}{2} = (\frac{3}{2} )$

Hence, f(0) =3/2 is TRUE.

c) Putting x = 1:

$f(1 ) = (\frac{1}{2})(1 ) +\frac{3}{2} = \frac{1}{2} + (\frac{3}{2} )\\\implies f(x) = \frac{3 + 1}{2} = (\frac{4}{2} ) = 2\implies 2 \neq -1$

Hence, f(1) = -1 is NOT TRUE.

d)Putting x = 2:

$f(2 ) = (\frac{1}{2})(2 ) +\frac{3}{2} = 1+ (\frac{3}{2} )\\\implies f(x) = \frac{2 + 3}{2} = (\frac{5}{2} )$

and (5/2) ≠ 1

Hence, f(2) = 1 is NOT TRUE.

e)Putting x = 4:

$f(4 ) = (\frac{1}{2})(4 ) +\frac{3}{2} = 2 + (\frac{3}{2} )\\\implies f(x) = \frac{4 + 3}{2} = (\frac{7}{2} )$

Hence, f(4) = 7/2 is TRUE.

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3 years ago

2x2-3x-2=x+4 select all the solutions

Papessa [141]

Solve for x over the real numbers:2 x^2 - 3 x - 2 = x + 4
Subtract x + 4 from both sides:2 x^2 - 4 x - 6 = 0
The left hand side factors into a product with three terms:2 (x - 3) (x + 1) = 0
Divide both sides by 2:(x - 3) (x + 1) = 0
Split into two equations:x - 3 = 0 or x + 1 = 0
Add 3 to both sides:x = 3 or x + 1 = 0
Subtract 1 from both sides:Answer: x = 3 or x = -1

5 0

3 years ago

Factor by grouping step by step 7c^4-4c^3+28c^2-16c

matrenka [14]

After factor by grouping $7c^4-4c^3+28c^2-16c$ we get $c(7c-4)(c^2+4)$