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vampirchik [111]

2 years ago

10

On a number line, the coordinates of X, Y, Z, and W are negative 7, negative 4, 3, and 7, respectively. Find the lengths

of the two segments below. Then tell whether they are congruent.
XY overbar and ZW overbar

PLEASE HELP ME

1 answer:

Vinil7 [7]2 years ago

6 0

Djdndisinx83 Jdjeijb

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Please help me with NUMBER 7 !! Thank you so much! 20 points!! :))

mart [117]

Answer:

x<-3 or x>-1

Step by step explanation:

Start with -5x>15. To get x by itself, we have to divide each side by -5. When we do that we get x<-3. (Don't forget to flip the sign when dealing with negative · and ÷.) Now answer the second equation. x-5>-6 To get x by itself, add 5 to both sides to get x>-1. Now we have x<3 or x>-1.

8 0

3 years ago

Factorization of:

zubka84 [21]

Answer:

Step-by-step explanation:

2800: 2 * 2 * 2 * 2 * 5 * 5 * 7

75: 3 * 5 * 5

168: 2 * 2* 2 * 3 * 7

It's not really multiplication. It's more division.

Try 2800 as a sample. What you are trying to do is break this down into primes. The first prime is 2

2800/2 = 1400

1400 / 2 = 700

700 / 2 = 350

350 / 2 = 175. That's the end of what the 2s can do.

175 / 5 = 35

35/ 5 = 7 7 is a prime. You are done. Now run up the ladder.

2800: 2 * 2 * 2 * 2 * 5 * 5 * 7

75 is not an even number. It has no 2s. Go to 3

75 / 3 = 25.

25 / 5 = 5

That's the end

75: 3 * 5 * 5

Your calculator can be of great help. The rule is keep factoring until you get a decimal remainder. Move on to the next prime. Stop when the last division gives you a prime.

3 0

3 years ago

The coffee pub has cans of coffee that weigh 4 4/5 pounds each. The pub has 8 1/2 cans of coffee left. What is the total weight?

tankabanditka [31]

4.8 x 8.5 = 40.8 lbs total
quick mafs

6 0

2 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago

Explain way in your mind to remember 9x6 and 8x7

d1i1m1o1n [39]

Add 9, 6 times
And add 8, 7 times

5 0

3 years ago

Read 2 more answers