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2 years ago

PLEASE SOLVE FOR X AND Y!!!! I WILL MARK BRAINLIEST!!!!

Mathematics

1 answer:

natima [27]2 years ago

4 0

opposite sides and diagonals of a parallelogram are equal

+ diagonals of a parallelogram meet eachother in the middle so:

10x+14=x+77

10x-x=77-14

9x=63

x=63÷9

x= 7

y+8=4y-19

y-4y=-19-8

-3y=-27

y=-27÷-3

y=9

2z+3=37

2z=37-3

2z=34

z=34÷2

z=17

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You are the manager of a local charity. You expect 63 donors to begin donating electronically next year. This figure

cluponka [151]

Answer:

237

Step-by-step explanation:

63 / .21 = 300

300 total people on list

21 percent are electronic so 79 percent are not electronic

300 x .79 = 237

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3 years ago

Factor x^2 + 4x + 340

iris [78.8K]

Answer:

its an unsolvible problem not kiding

Step-by-step explanation:

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4 years ago

Kelly walks .7 miles to school Mary walks .49 miles to school write an equation using greater than less than or equal to to comp

kupik [55]

Kelly .7 > Mary .49
Because kelly is actually .70 away from school when Mary is .49.
So Kelly would be farther away from school then Mary.

8 0

3 years ago

Read 2 more answers

Simplify the radical expression. 2 sqrt6 + 3 sqrt 96

Ksenya-84 [330]

The answer would be a) 14 sqrt 6

6 0

3 years ago

Read 2 more answers

The initial size of the population is 300. After 1 day the population has grown to 800. Estimate the population after 6 days. (R

Cloud [144]

Solution :

Given initial population = 300

Final population after 1 day = 800

Number of days = 6

∴ $\frac{dP}{dt} =kt^{1/2}$

P(0) = 300 P(1) = 300

We need to find P(8).

$dP = kt^{1/2} dt$

$\int 1 dP = \int kt^{1/2} dt$

$P(t) = k \left(\frac{t^{3/2}}{3/2}\right)+c$

$P(t)= \frac{2k}{3}t^{3/2} + c$

When P(0) = 300

$300 = \frac{2k}{3} (0)^{3/2} + c$

∴ c = 300

∴ $P(t)= \frac{2k}{3}t^{3/2} + 300$

When P(1) = 800

$800 = \frac{2k}{3} (1)^{3/2} + 300$

$500 = \frac{2k}{3}$

∴ k = 750

$P(t)= 500t^{3/2} + 300$

So, P(8) is

$P(t)= 500(8)^{3/2} + 300$

= 11,614

So the population becomes 11,614 after 8 days.

8 0

3 years ago