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3 years ago

15a + 22b = 540a + 55b = 13Can someone please please help

Mathematics

2 answers:

Vlada [557]3 years ago

6 0

Answer:

value of a is 1/5 and b=1/11

Mazyrski [523]3 years ago

5 0

Https://photomath.net/s/arvkdM

Use Photomath app to get the steps

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Solve for x. SHOW ALL WORK FOR CREDIT! ( 35pts)

iren2701 [21]

Answer:

Step-by-step explanation:

they are two opposite angles, therefore equal

14x - 4 = 13x + 2

14x - 13x = 2 + 4

x = 6

-----------------

check

14 * 6 - 4 = 13 * 6 + 2

80 = 80

<em>the answer is good</em>

8 0

2 years ago

Find the quotient. 48a 3 bc 2 ÷ 3abc

adoni [48]

The answer is 96.

The 3,a,b,& c cancel out because they equal 1.
Then your left with 48*2 which is 96.

48 * a * 3 * b * c * 2--------------------------3 * a * b * c
3/3 = 1 B/b = 1 C/c = 1
1 * 1 * 1 * 48 * 2 = 96

3 0

4 years ago

Read 2 more answers

QUESTION 2

aev [14]

The margin of error for a 95% confidence interval is 0.93.

mean (μ) = 23.4, standard deviation (σ) = 7.6, sample (n) = 254, confidence = 95% = 0.95

α = 1 - C = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

The z score of α/2 is equal to the z score of 0.475 (0.5 - 0.025) which is equal to 1.96.

The margin of error (E) is given by:

$E=z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } \\\\E=1.96*\frac{7.6}{\sqrt{254} } \\\\E=0.93$

The margin of error for a 95% confidence interval is 0.93.

Find out more at: brainly.com/question/10501147

4 0

3 years ago

How do u go pro at skateboarding? Also why is my baseball cup so uncomfortable?

dimulka [17.4K]

Answer:

You can probably go pro at skating by just practicing a ton and learning new skills and tricks that separates you from other skaters. Its not very hard to be a "pro", its hard to make money off of skating because of the lack of audience it has. Also, I assume your baseball cap is uncomfortable because of the material or your hair.

Hope that helps

8 0

3 years ago

Find the area of the surface. The part of the sphere x2 + y2 + z2 = a2 that lies within the cylinder x2 + y2 = ax and above the

sineoko [7]

Answer:

The area of the sphere in the cylinder and which locate above the xy plane is $\mathbf{ a^2 ( \pi -2)}$

Step-by-step explanation:

The surface area of the sphere is:

$\int \int \limits _ D \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 ) } \ dA$

and the cylinder $x^2 + y^2 =ax$ can be written as:

$r^2 = arcos \theta$

$r = a cos \theta$

where;

D = domain of integration which spans between $\{(r, \theta)| - \dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}, 0 \leq r \leq acos \theta\}$

and;

the part of the sphere:

$x^2 + y^2 + z^2 = a^2$

making z the subject of the formula, then :

$z = \sqrt{a^2 - (x^2 +y^2)}$

Thus,

$\dfrac{\partial z}{\partial x} = \dfrac{-2x}{2 \sqrt{a^2 - (x^2+y^2)}}$

$\dfrac{\partial z}{\partial x} = \dfrac{-x}{ \sqrt{a^2 - (x^2+y^2)}}$

Similarly;

$\dfrac{\partial z}{\partial y} = \dfrac{-2y}{2 \sqrt{a^2 - (x^2+y^2)}}$

$\dfrac{\partial z}{\partial y} = \dfrac{-y}{ \sqrt{a^2 - (x^2+y^2)}}$

So;

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\begin {pmatrix} \dfrac{-x}{\sqrt{a^2 -(x^2+y^2)}} \end {pmatrix}^2 + \begin {pmatrix} \dfrac{-y}{\sqrt{a^2 - (x^2+y^2)}} \end {pmatrix}^2+1}$ $\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{x^2+y^2}{a^2 -(x^2+y^2)}+1}$

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{x^2+y^2+a^2 -(x^2+y^2)}{a^2 -(x^2+y^2)}}$

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{a^2}{a^2 -(x^2+y^2)}}$

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = {\dfrac{a}{\sqrt{a^2 -(x^2+y^2)}}$

From cylindrical coordinates; we have:

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = {\dfrac{a}{\sqrt{a^2 -r^2}}$

dA = rdrdθ

By applying the symmetry in the x-axis, the area of the surface will be:

$A = \int \int _D \sqrt{ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2+1} \ dA$

$A = \int^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}} \int ^{a cos \theta}_{0} \dfrac{a}{\sqrt{a^2 -r^2 }} \ rdrd \theta$

$A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 -r^2} \end {bmatrix}^{a cos \theta}_0 \ d \theta$

$A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 - a^2cos^2 \theta} + a \sqrt{a^2 -0}} \end {bmatrix} d \theta$ $A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \ sin \theta +a^2 } \end {bmatrix} d \theta$