*HELP*
calculate the pay for the following day of a weekly time card given a wage of $12 an hour

Answer:

$420

.8 x = 336

x = 336/.8

X=$420

Step-by-step explanation:

Answer:

height of tree is 27 ft

Step-by-step explanation:

The corresponding parts of the problem are in proportion

let h be the height of the tree , then

[tex]\frac{6}{h}[/tex] = [tex]\frac{7}{31.5}[/tex] ( cross- multiply )

7h = 189 ( divide both sides by 7 )

h = 27

Height of tree is 27 feet

Answer:

a, a, d

Step-by-step explanation:

44

The area (A) of a triangle is calculated as

A = [tex]\frac{1}{2}[/tex] bh ( b is the base and h the perpendicular height )

Here b = 10 and h = 9 , then

A = [tex]\frac{1}{2}[/tex] × 10 × 9 = 5 × 9 = 45 in² → a

45

The area (A) of a parallelogram is

A = bh ( b is the base and h the perpendicular height )

Here b = 2 and h = 4 , then

A = 2 × 4 = 8 m² → a

46

A = bh ( with b = 4 and h = 10 )

A = 4 × 10 = 40 m² → d

Answer

25mm of rain fell each second

Step-by-step explanation:

we know that 500 mm fell in 20 minutes

so, we have to divide 500 by 20 giving us the amount of rain that fell each minute:

500/20 = 25

therefore, 25mm of rain fell each minute

hope this helped:)

Answer:

25 mm per minute

Step-by-step explanation:

Take the amount of rain and divide by the number of minutes

500/20

25 mm per minute

Answer:

C. 7 units

Step-by-step explanation:

The given parameters are;

The length of the chord of the circle, [tex]\overline{AC}[/tex] = 14 units

The orientation of the radius and the chord = The radius is perpendicular to the chord

We have in ΔAOC, [tex]\overline{AO}[/tex] = [tex]\overline{OC}[/tex] = The radius of the circle

[tex]\overline{OB}[/tex] ≅ [tex]\overline{OB}[/tex]  by reflexive property

The angle at point B = 90° by angle formed by the radius which is perpendiclar to the chord [tex]\overline{AC}[/tex]

ΔAOB and ΔCOB are right triangles (triangles having one 90° angle)

[tex]\overline{AO}[/tex] and [tex]\overline{OC}[/tex] are hypotenuse sides of ΔAOB and ΔCOB respectively and [tex]\overline{OB}[/tex] is a leg to ΔAOB and ΔCOB

Therefore;

ΔAOB ≅ ΔCOB, by Hypotenuse Leg rule of congruency

Therefore;

[tex]\overline{AB}[/tex] ≅ [tex]\overline{BC}[/tex] by Congruent Parts of Congruent Triangles are Congruent, CPCTC

[tex]\overline{AB}[/tex] = [tex]\overline{BC}[/tex] by definition of congruency

[tex]\overline{AC}[/tex] = [tex]\overline{AB}[/tex] + [tex]\overline{BC}[/tex] by segment addition postulate

∴ [tex]\overline{AC}[/tex] = [tex]\overline{AB}[/tex] + [tex]\overline{BC}[/tex] =  [tex]\overline{AB}[/tex] + [tex]\overline{AB}[/tex] = 2 ×  [tex]\overline{AB}[/tex]

∴  [tex]\overline{AB}[/tex] = [tex]\overline{AC}[/tex]/2

[tex]\overline{AB}[/tex] = 14/2 = 7

[tex]\overline{AB}[/tex] = 7 units.

Answer:

7 units

Step-by-step explanation:

Answer:

here's the answer to your question

Answer: √18

√(4-1)^2 + (5-2)^2

√9 + 9

√18

Answered by Gauthmath must click thanks and mark brainliest

Answer/Step-by-step explanation:

1. To find the area of the shaded region, you'd find the area of the white rectangular shape, next, find the area of the whole triangular shape, then find the difference of their areas to get the area of the shaded region. Thus, the formula to use would be:

Area of shaded region = area of triangle - area of rectangle

Area of shaded region = ½*base*height - length*width

1. a. Volume of triangular prism = area of triangular base * height of prism

Volume of triangular prism = ½bh * H

Where,

b = 6 m

h = 4 m

H = 8 m

Substitute

Volume of prism = ½*6*4*8

Volume of prism = 96 m³

b. Volume of sphere = ⁴/3πr³

Where,

r = 9 cm

Substitute

Volume = ⁴/3*π*9³

Volume = ⁴/3*π*729

Volume ≈ 3,053.6 cm³ (nearest tenth)

2. Use Pythagorean theorem to find the height of the cone

radius of the cone (r) = ½(16) = 8 cm

Slant height (l) = 11 cm

height (h) = ?

Using Pythagorean theorem, we have:

h = √(l² - r²)

Substitute

h = √(11² - 8²)

h = √(57)

h ≈ 7.5 cm (nearest tenth)

b. Volume of the cone = ⅓πr²h

where,

r = 8 cm

h = 7.5 cm

Volume = ⅓*π*8²*7.5

Volume = 502.7 cm³ (nearest tenth)

Answer:

d no.

Step-by-step explanation:

because in my my personal opinion the function shown on the graph has a smaller rate of change, but a higher starting point

Answer:

18[tex]\pi[/tex]

[tex]\frac{80}{360} * 81 \pi[/tex]

Step-by-step explanation:

Answer:

C: 119 square inches

Step-by-step explanation:

We are given;

Slant height; L = 5 in

Base area; B = 49 in²

Since it's a square pyramid, the base portion has a square shape.

Thus, area of base = x²

Where x is a side of the square.

Thus;

x² = 49

x = √49

x = 7

Perimeter of base = 7 × 4 = 28 in

Area of pyramid = ½PL + B

Plugging in the relevant values;

Area of pyramid = (½ × 28 × 5) + 49

Area of pyramid = 119 in²

Answer:

A.1/3 B.5/8 C.1/25

Step-by-step explanation:

just simplify it down to the lowest

Answer:

Ä. 16/24 as reduced fraction = 2/3

B. 15/24 as reduced fraction= 5/8

Answer:

0.63 ÷ 0.3 = 63 x 10^-2 ÷ 3 x 10^-1 = 63 / 3  x 10^(-2--1) = 21 x 10^-1 = 2.1

Answer:

2.1.

Step-by-step explanation:

63/3 = 21

0.63 = 63 / 100

and 0.3 = 3 / 10

so 0.63/0.3 = 63 /100 / 3/10 = 63/100 * 10/3

= 63/3 / 10

So the answer is 21 / 10 = 2.1.

Answer:

27/20

Step-by-step explanation:

[tex]\frac{13.5}{10} = \frac{13.5 * 2}{10 * 2} = \frac{27}{20}[/tex]

Answer:

Step-by-step explanation:

The formula for slope is y2-y1/x2-x1 where y2 and x2 are the x and y coordinates from a coordinate pair and y1 and x1 are the coordinates from another coordinate pair. In this case, 2 coordinate pairs are given: (30,75) and (10, 35)  75-35/30-10 would be your slope, or, 40/20, or simplified, 2.

Your slope is 2

Answer:

y = 23, -31

Step-by-step explanation

ttyl

Answer:

A.

Step-by-step explanation:

the square root of 6 is roughly 1.57

so that means the ordered pair would read (2,1.57).

if you were to plot that point it would be on the circle.

Also the distance from the origin (0,0) to (3,0) is 3 units

We will see that the correct option is:

"No, the distance from (0, 0) to (2, 6) is not 3 units."

We know that the circle has a radius of 3 units, then we need to see if the distance between (0, 0) and (2, 6) is 3 units.

Here we have:

[tex]D = \sqrt{(6 - 0)^2 + (2 - 0)^2} = \sqrt{36 + 4} = \sqrt{40} \neq 3[/tex]

So the distance between (0, 0) and (2, 6) is different than 3 units, meaning that the point is not in the circle.

If you want to learn more about circles:

https://brainly.com/question/1559324

#SPJ5

Answer:

x = 2

Step-by-step explanation:

Answer:

There are 0.454 kg in one pound.

So, in 120 pounds there are 0.454 x 120 kgs.

This is equal to 54.48, and the answer is 54.48 kg.

Let me know if this helps!

Answer:

7

Step-by-step explanation:

Going from -5 to 2 we get

1) -4

2) -3

3) -2

4) -1

5) 0

6) 1

7) 2

So, in total, there are 7 numbers between -5 and 2

Answer:

where are the numbers

Step-by-step explanation:

Answer: False

The variable portions x^2y and xy^2 are slightly different, so that's why we don't have like terms here.

Answer:

False

Step-by-step explanation:

Like terms are those terms where they have the same variable

so here

5x2y variables are = 5x and 2y

7xy2 variables are = 7x and y2

you can observe that x variable is same but y variable is not same as 5x3y have 2y whereas 7xy2 have y2#

so the answer is FALSE

Answer:

i hope it will help

Step-by-step explanation:

I did not get the equation so I solve it with two methods

Answer:

Step-by-step explanation:

For Part A, use the pythagorean theorem to find the height, which ca be found by finding one length of the leg. Using the imaginary bisector, you can determine one of the legs is 5 cm, and the hypotenuse is 13 cm

a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse. Plug in the values to solve for one leg, you get 25+b^2=169

Solve algebraically, b^2= 144, so b=12, which is the height.

Part B

Determine the surface area of each cardboard piece, and add together.

20 × 13 × 2 = 520

1/2 × 10 × 12 × 2 = 120

20 × 10 = 200

So approximately 840 cm of cardboard was used

Answer:

a= [tex]\sqrt{19}[/tex]

Step-by-step explanation:

Pythagorean Theorem: a^2+b^2=c^2

a^2+9^2=10^2

a^2+81=100

a^2=19

a=[tex]\sqrt{19}[/tex]

Answer:

b = 4.4 meters

Answer: 79

Concept:

Here, we need to understand the idea of evaluation.  

When encountering questions that gave you an expression with variables, then stated: "If x = a, y = b, z = c" (a, b, c are all constants), this means you should substitute the value given for each variable back to the expression.

Solve:

Given information

10x² - 7x + 10

x = 3

Substitute the value into the expression

= 10 (3)² - 7 (3) + 10

Simplify by multiplication

= 10 (9) - 21 + 10

= 90 - 21 + 10

Simplify by subtraction

= 69 + 10

Simplify by addition

= 79

Hope this helps!! :)

Please let me know if you have any questions

Answer:

92

Step-by-step explanation:

The variable 'x' shows up twice in this expression.  Replace each instance of 'x' with 3:

10(3)^2 - 7(3) + 10  =  10(9) - 21 + 19  =  92

Answer:

[tex]a. \ \dfrac{625 \cdot m}{27 \cdot n^{11}}[/tex]

[tex]b. \ \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}[/tex]

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

[tex]\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}[/tex]

By expanding the expression, we get;

[tex]\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}[/tex]

Collecting like terms gives;

[tex]\dfrac{m^{(3 + 4 - 6)} \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}[/tex]

[tex]\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}[/tex]

(b) The given expression is presented as follows;

[tex]x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4[/tex]

Therefore, we get;

[tex]x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n}[/tex]

Collecting like terms gives;

[tex]x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )[/tex]

[tex]x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}[/tex]

[tex]x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}[/tex]