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3 years ago

6

Can someone plz help me with this one problem plzzzzz!!! I’M MARKING BRAINLIEST!!!!

1 answer:

marishachu [46]3 years ago

7 0

Answer:

4-1=3

6-1=5

8-1=7

Step-by-step explanation:

With c= w-1 all you have to do is plug in your answers and solve for the rest. Hope it helps!

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Use the elimination method to find all solutions of the system:

weqwewe [10]

Answer:

$x = 3$ and $y = 4$

Step-by-step explanation:

Given

[Correct Question]

$x^2 - 2y = 1$

$x^2 + 5y = 29$

Required

Solve using elimination

First, we eliminate x.

Subtract equation (1) from (2)

$x^2 - x^2 + 5y - (-2y) = 29 - 1\\$

$5y - (-2y) = 28$

$5y +2y = 28$

$7y = 28$

Divide through ny 7

$y = 4$

Take $x^2 - 2y = 1$

Make x^2 the subject

$x^2 = 1 + 2y$

Substitute 4 for y

$x^2 = 1 + 2*4$

$x^2 = 1 + 8$

$x^2 = 9$

Take square roots

$x = \sqrt 9$

$x = 3$

7 0

2 years ago

How do you write 9,800,100 in exponents and expanded form

Leokris [45]

In exponents it could be 3130.51114^2, 213.998224 and 55.95097086
Happy to Help!

5 0

3 years ago

Suppose that an insect population’s density, in thousands per acre, during year n, can be modeled by the recursive formula: a1 =

vlabodo [156]

Answer:

The population alternates between increasing and decreasing

Step-by-step explanation:

<u><em>The options of the question are</em></u>

A) The population density decreases each year.

B) The population density increases each year.

C) The population density remains constant.

D) The population alternates between increasing and decreasing

we have

$a_n=2.9(a_n_-_1)-0.2(a_n_-_1)^2$

$a_1=8$

Find the value of $a_2$

For n=2

$a_2=2.9(a_1)-0.2(a_1)^2$

$a_2=2.9(8)-0.2(8)^2=10.4$

Find the value of $a_3$

For n=3

$a_3=2.9(a_2)-0.2(a_2)^2$

$a_3=2.9(10.4)-0.2(10.4)^2=8.528$

For n=4

$a_4=2.9(a_3)-0.2(a_3)^2$

$a_4=2.9(8.528)-0.2(8.528)^2=10.1858$

For n=5

$a_5=2.9(a_4)-0.2(a_4)^2$

$a_5=2.9(10.1858)-0.2(10.1858)^2=8.7887$

therefore

The population alternates between increasing and decreasing

7 0

3 years ago

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nexus9112 [7]

U need upload the image

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2 years ago

pav-90 [236]

20-24 ...................

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3 years ago