Write an expression for "the quotient of z and 4."

Find the surface area of this prism please (no links please)

horsena [70]

Answer:

192 ft^2

Step-by-step explanation:

to find the surface area of a shape, you must find the area of each side and add them together. the base of this shape is a rectangle with a length of 11ft and a height of 3ft; to calculate the area of a rectangle, you multiply the length by the height. 3*11 is 33, so the base of this prism has an area of 33ft^2. since this is a prism, we know that the side parallel to the base are the same shape, so they both have an area of 33ft^2. 33*2= 66. the two other large sides have a height of 4.5 and a length of 11; 4.5*11 is 49.5, there are two shapes with these dimensions, so 49.5*2=99. finally, the two shapes on either end of the prism have a height of 4.5 and a length of 3. 4.5*3= an area of 13.5, two of them, so 13.5*2=27. now you just add all of these numbers together: 66+99+27=192.

4 0

3 years ago

Someone please help me with the question above. It uses the words ‘income tax’ and I have a hard time understanding the differen

lesya [120]

Step-by-step explanation:

incone tax means the money u get as income from tax and tax payable is the tax needed to be to payed

6 0

3 years ago

The sum of 2 numbers (x and y) is 20, and their product is 51. What are the values of x and y

Murljashka [212]

Answer: 13&7

Step-by-step explanation:

The sum of two numbers x&y is 20

Product of the 2numbers xy is 51

X+y=20......equation 1

XY=51.........equation 2

X=20-y.......equation 3

Substitute for x inequation 3 into equation 2

So we have;

(20-y)y=51

Open the bracket

20y-y^2=51

Collect like terms,so we have

Y2-20y+51=0

Factorise

Factors of 51 is 13&7

So -13-7=-20 &-13×-7=51

Substitute

Y2-13y-7y+51=0

Y(y-13)-7(y-13)=0

(Y-7)(y-13)=0

Y-7=0

Y=7 or

Y-13=0

Y=13

Y=13 or 7

Substitute for y in equation 1

X+y=20

X+7=20

X=20-7

X=13 or

X+13=20

X=20-13

X=7

Therefore, X &Y = 13 &7

4 0

3 years ago

Unit 8 right triangles and trigonometry homework 3 similar right triangles and geometric mean

Ugo [173]

The right triangles that have an altitude which forms two right triangles

are similar to the two right triangles formed.

Responses:

1. ΔLJK ~ ΔKJM

ΔLJK ~ ΔLKM

ΔKJM ~ ΔLKM

2. ΔYWZ ~ ΔZWX

ΔYWZ ~ ΔYZW

ΔZWX ~ ΔYZW

3. x = 4.8

4. x ≈ 14.48

5. x ≈ 11.37

6. G.M. = 12·√3

7. G.M. = 6·√5

<h3>What condition guarantees the similarity of the right triangles?</h3>

1. ∠LMK = 90° given

∠JMK + ∠LMK = 180° linear pair angles

∠JMK = 180° - 90° = 90°

∠JKL ≅ ∠JMK All 90° angles are congruent

∠LJK ≅ ∠LJK reflexive property

ΔLJK is similar to ΔKJM by Angle–Angle, AA, similarity postulate

∠JLK ≅ ∠JLK by reflexive property

ΔLJK is similar to ΔLKM by AA similarity

By the property of equality for triangles that have equal interior angles, we have;

ΔKJM ~ ΔLKM

2. ∠YWZ ≅ ∠YWZ by reflexive property

∠WXZ ≅ ∠YZW all 90° angle are congruent

ΔYWZ is similar to ΔZWX, by AA similarity postulate

∠XYZ ≅ ∠WYZ by reflexive property

∠YXZ ≅ ∠YZW all 90° are congruent

ΔYWZ is similar to ΔYZW by AA similarity postulate

Therefore;

ΔZWX ~ ΔYZW

3. The ratio of corresponding sides in similar triangles are equal

From the similar triangles, we have;

$\dfrac{8}{10} = \mathbf{ \dfrac{x}{6}}$

8 × 6 = 10 × x

48 = 10·x

$x = \dfrac{48}{10} = \underline{4.8}$

3. From the similar triangles, we have;

$\mathbf{\dfrac{20}{29}} = \dfrac{x}{21}$

20 × 21 = x × 29

420 = 29·x

$x = \dfrac{420}{29 } \approx \underline{14.48}$

4. From the similar triangles, we have;

$\mathbf{\dfrac{20}{52}} = \dfrac{x}{48}$

20 × 48 = 52 × x

$x = \dfrac{20 \times 48}{52} = \dfrac{240}{13} \approx \underline{18.46}$

5. From the similar triangles, we have;

$\mathbf{\dfrac{13.2}{26}} = \dfrac{x}{22.4}$

13.2 × 22.4 = 26 × x

$x = \dfrac{13.2 \times 22.4}{26} \approx \underline{ 11.37}$

6. The geometric mean, G.M. is given by the formula;

$G.M. = \mathbf{\sqrt[n]{x_1 \times x_2 \times x_3 ... x_n}}$

The geometric mean of 16 and 27 is therefore;

$G.M. = \sqrt[2]{16 \times 27} = \sqrt[2]{432} = \sqrt[2]{144 \times 3} = \mathbf{12 \cdot \sqrt{3}}$

The geometric mean of 16 and 27 is 12·√3

7. The geometric mean of 5 and 36 is found as follows;

$G.M. = \sqrt[2]{5 \times 36} = \sqrt[2]{180} = \sqrt[2]{36 \times 5} = \mathbf{ 6 \cdot \sqrt{5}}$

The geometric mean of 5 and 36 is 6·√5

Learn more about the AA similarity postulate and geometric mean here:

brainly.com/question/12002948

brainly.com/question/12457640

7 0

3 years ago

If you were asked to solve a system of linear equations using the substitution method, in what form would one or both of the equ

elena55 [62]

Answer:

It'd be easist if one of the variables was by itself

Step-by-step explanation:

I.E: y=2x+4 and 2y+x=23

4 0

4 years ago

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