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3 years ago

Solve for z −4z + 31 ≥ 17z + 23

Mathematics

2 answers:

mojhsa [17]3 years ago

6 0

Answer:

z ≤ 8/21

Step-by-step explanation:

Flip the inequality sign since you are multiplying by a negative number

21z-8 ≤ 0

1.2 Divide both sides by 21

z-(8/21) ≤ 0

Add 8/21 to both sides

z ≤ 8/21

Natali [406]3 years ago

3 0

$z \leqslant \frac{8}{21}$

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Step-by-step explanation:

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3 years ago

A graduated commission employee makes 3.5% interest on the first $50,000 in sales and 6.5% interest on all sales over $50,000. W

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Step-by-step explanation:

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3 years ago

What is the explicit rule for the sequence? 12,6,0,−6, …

wolverine [178]

Hello!

First you find the common difference in the sequence

6 - 12 = -6
0 - 6 = -6

The common difference is -6

We can now make the equation

-6(n - 1) + 12

we got the 12 because it is the first number in the sequence

Simplify

-6n + 6 + 12

Simplify

18 - 6n

The answer is B) an = 18 - 6n

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7 0

3 years ago

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo

oksano4ka [1.4K]

Answer:

Required center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

Step-by-step explanation:

Given semcircles are,

$y=\sqrt{1-x^2}, y=\sqrt{16-x^2}$ whose radious are 1 and 4 respectively.

To find center of mass, $(\bar{x},\bar{y})$ , let density at any point is $\rho$ and distance from the origin is r be such that,

$\rho=\frac{k}{r}$ where k is a constant.

Mass of the lamina=m= $\int\int_{D}\rho dA$ where A is the total region and D is curves.

then,

$m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k$

Now, x-coordinate of center of mass is $\bar{y}=\frac{M_x}{m}$ . in polar coordinate $y=r\sin\theta$

$\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta$

$=3k\int_{0}^{\pi}\sin\theta d\theta$

$=3k\big[-\cos\theta\big]_{0}^{\pi}$

$=3k\big[-\cos\pi+\cos 0\big]$

$=6k$

Then, $\bar{y}=\frac{M_x}{m}=\frac{2}{\pi}$

y-coordinate of center of mass is $\bar{x}=\frac{M_y}{m}$ . in polar coordinate $x=r\cos\theta$

$\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta$

$=3k\int_{0}^{\pi}\cos\theta d\theta$

$=3k\big[\sin\theta\big]_{0}^{\pi}$

$=3k\big[\sin\pi-\sin 0\big]$

$=0$

Then, $\bar{x}=\frac{M_y}{m}=0$

Hence center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

3 0

3 years ago

molly school is selling tickets to the fall musical. on the first day of the ticket sales the school sold 13 adults tickets and

tester [92]

Answer:

Adult Ticket Price = $13

Child Ticket Price = $13

Step-by-step explanation:

Let price of adult ticket be "a" and child ticket be "c"

13 adult and 14 child equals $351, so we can write:

13a + 14c = 351

and

2 adult and 7 child equals $117, thus we can write:

2a + 7c = 117

We multiply 2nd equation by (-2) to get:

-2 * [2a + 7c = 117]

= -4a -14c = -234

Adding botht he "bold" equations, we get:

13a + 14c = 351

-4a -14c = -234

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

9a = 117

a = 117/9 = 13

Now to find b, we use the value of a gotten in the first equation:

13a + 14c = 351

13(13) + 14c = 351

169 + 14c = 351

14c = 182

c = 182/14 = 13

Hence,

Adult Ticket Price = $13

Child Ticket Price = $13

3 0

3 years ago