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Andrew [12]
2 years ago
9

Please see attached image and help with these absolute value problems - thank you

Mathematics
1 answer:
Novay_Z [31]2 years ago
6 0

Answer:

{-infinity, infinity}

no solution

Step-by-step explanation:

|x| +15≥4

Subtract 15 from each side

|x| +15-15≥4-15

|x| ≥-11

This is always true so x can be and real number

{-infinity, infinity}

-2|5x+3| = 8

Divide by -2

-2/-2|5x+3| = 8/-2

|5x+3| = -4

This can never happen because absolute values are zero or greater

no solution

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Math ... answer meeee
zloy xaker [14]

You need to find the GCF (greatest common factor) of 45 and 90, which is 45.

45 times ? gets you 45

? would be 1

45 times 1 equals 45.

45 times ? gets you 90

? would be 2

45 times 2 equals 90

45(1+2) = 45+90 = 135

♡ Hope this helps! ♡

❀ 0ranges ❀

7 0
3 years ago
Read 2 more answers
9x + 26 + 7x - 17 = 2x + (-3x) + 5x<br><br> 16x + 9 = 0<br> 16x + 11 = 4x<br> 16x + 9 = 4x
zzz [600]

Answer:

C. 16x+9=4x

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
The cost of a school banquet is 75$ plus 15$ per person.
ivann1987 [24]

Answer:

The cost of a school banquet is $75+30n, where n is the number of people attending.

If 53 people are attending to visit school banquet, then you have to find the value of given expression at n=53.

At n=53, the value is $(75+30·53)=$1,665.

Step-by-step explanation:

4 0
2 years ago
Jacob is 5 5/6 feet tall. Linda is 5 1/4 feet tall. How much taller is Jacob?
SCORPION-xisa [38]
This problem boils down to 5/6 - 1/4

To subtract these fractions you need a common denominator.

I'll use 12

The first fraction becomes 10/12 and the second fraction becomes 3/12

10/12-3/12 = 7/12

So Jacob is 7/12 of a foot taller than Linda.

Also note than since 12 inches = 1 foot, you could say Jacob is 7 inches taller than Linda
6 0
3 years ago
Read 2 more answers
Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=
Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = x^2 + y^2 +z^2 = 4^2

z^2 = 16 -x^2 -y^2

z = \sqrt{16-x^2-y^2}

The partial derivatives of Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}

Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}

Similarly;

Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}

∴

dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA

=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA

=\sqrt{ \dfrac{16}{16-x^2-y^2}  }\ \ .dA

=\dfrac{4}{\sqrt{ 16-x^2-y^2}  }\ \ .dA

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to \sqrt{12}

dA = rdrdθ

∴

The surface area S = \int \limits _R \int \ dS

=  \int \limits _R\int  \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA

= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \  \ rdrd \theta

= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr

= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}

= 8\pi ( - \sqrt{4} + \sqrt{16})

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0
2 years ago
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