Find the solution to the following system using substitution or elimination:

Find x 25 7 х I need help

Murljashka [212]

Answer:

24 = x

Step-by-step explanation:

You use Pythagorean theorem. Since the normal equation is $a^{2} + b^{2} =c^{2}$ , you already have the $c^{2}$ and another side which can be $a^{2}$ or $b^{2}$ .

Because you have the hypotenuse and a side , your equation looks like this now: $7^{2} +b^{2} = 25^{2}$

and when you solve for be you just subtract $25^{2} -7^{2}$ to get 24.

8 0

2 years ago

Can anyone plz help me

ser-zykov [4K]

Answer:

C. Veggie Delight

Step-by-step explanation:

1/5 is 20 cents so 40 cents make a pound then you compare them to the rest.

6 0

3 years ago

Read 2 more answers

Help me on this please

zalisa [80]

Answer:

1. (x, y) → (x + 3, y - 2)

Vertices of the image

a) (-2, - 3)

b) (-2, 3)

c) (2, 2)

2. (x, y) → (x - 3, y + 5)

Vertices of the image

a) (-3, 2)

b) (0, 2)

c) (0, 4)

d) (2, 4)

3. (x, y) → (x + 4, y)

Vertices of the image

a) (-1, -2)

b) (1, -2)

c) (3, -2)

4. (x, y) → (x + 6, y + 1)

Vertices of the image

a) (1, -1)

b) (1, -2)

c) (2, -2)

d) (2, -4)

e) (3, -1)

f) (3, -3)

g) (4, -3)

h) (1, -4)

5. (x, y) → (x, y - 4)

Vertices of the image

a) (0, -2)

b) (0, -3)

c) (2, -2)

d) (2, -4)

6. (x, y) → (x - 1, y + 4)

Vertices of the image

a) (-5, 3)

b) (-5, -1)

c) (-3, 0)

d) (-3, -1)

Explanation:

To identify each IMAGE you should perform the following steps:

List the vertex points of the preimage (the original figure) as ordered pairs.

Apply the transformation rule to every point of the preimage

List the image of each vertex after applying each transformation, also as ordered pairs.

1. (x, y) → (x + 3, y - 2)

The rule means that every point of the preimage is translated three units to the right and 2 units down.

Vertices of the preimage Vertices of the image

a) (-5,2) (-5 + 3, -1 - 2) = (-2, - 3)

b) (-5, 5) (-5 + 3, 5 - 2) = (-2, 3)

c) (-1, 4) (-1 + 3, 4 - 2) = (2, 2)

2. (x,y) → (x - 3, y + 5)

The rule means that every point of the preimage is translated three units to the left and five units down.

Vertices of the preimage Vertices of the image

a) (0, -3) (0 - 3, -3 + 5) = (-3, 2)

b) (3, -3) (3 - 3, -3 + 5) = (0, 2)

c) (3, -1) (3 - 3, -1 + 5) = (0, 4)

d) (5, -1) (5 - 3, -1 + 5) = (2, 4)

3. (x, y) → (x + 4, y)

The rule represents a translation 4 units to the right.

Vertices of the preimage Vertices of the image

a) (-5, -2) (-5 + 4, -2) = (-1, -2)

b) (-3, -5) (-3 + 4, -2) = (1, -2)

c) (-1, -2) (-1 + 4, -2) = (3, -2)

4. (x, y) → (x + 6, y + 1)

Vertices of the preimage Vertices of the image

a) (-5, -2) (-5 + 6, -2 + 1) = (1, -1)

b) (-5, -3) (-5 + 6, -3 + 1) = (1, -2)

c) (-4, -3) (-4 + 6, -3 + 1) = (2, -2)

d) (-4, -5) (-4 + 6, -5 + 1) = (2, -4)

e) (-3, -2) (-3 + 6, -2 + 1) = (3, -1)

f) (-3, -4) (-3 + 6, -4 + 1) = (3, -3)

g) (-2, -4) (-2 + 6, -4 + 1) = (4, -3)

h) (-2, -5) (-2 + 3, -5 + 1) = (1, -4)

5. (x, y) → (x, y - 4)

This is a translation four units down

Vertices of the preimage Vertices of the image

a) (0, 2) (0, 2 - 4) = (0, -2)

b) (0,1) (0, 1 - 4) = (0, -3)

c) (2, 2) (2, 2 - 4) = (2, -2)

d) (2,0) (2, 0 - 4) = (2, -4)

6. (x, y) → (x - 1, y + 4)

This is a translation one unit to the left and four units up.

Vertices of the pre-image Vertices of the image

a) (-4, -1) (-4 - 1, -1 + 4) = (-5, 3)

b) (-4 - 5) (-4 - 1, -5 + 4) = (-5, -1)

c) (-2, -4) (- 2 - 1, -4 + 4) = (-3, 0)

d) (-2, -5) (-2 - 1, -5 + 4) = (-3, -1)

8 0

3 years ago

Jade has $451.89 in her checking account. After 3 checks, each for the same amount, are deducted from her account, she has $358.

Lana71 [14]

and the second one 2 will work!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

6 0

3 years ago

Read 2 more answers

Enter your answer in the provided box. what is the emf of a cell consisting of a pb2 / pb half-cell and a pt / h / h2 half-cell

s2008m [1.1K]

The emf of a cell is mathematically given as

$E_{cell}$ = 0.0532 v

Generally, the equation for emf is mathematically given as

$Pb/Pb^2^+ //\ H / H_2$

where,

Pb is lead

H is hydrogen

$Pb^{2+}$ / Pb has a standard reduction potential of -0.126 volts, while

$H_2$ / $H_2$ has a standard reduction potential of 0.0 volts.

$E_{cell }= E^0_{cell} - 0.059/n log 1/$ $[Pb^2^+] x [H^+]^2$

$E_{cell}$ = 0.126 - 0.059/2 log 1 / [0.57] x [0.09]^2

$E_{cell}$ = 0.0532 v