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3 years ago

Ms. Hartline grades one math test every 1⁄4 hour. How many math tests does she grade in 2 hours? HELP QUICK!!

Mathematics

1 answer:

vazorg [7]3 years ago

4 0

Answer:

8 maths tests

Step-by-step explanation:

Given data

We are told that in 1/4 hour that is 0.25hour (15 mins) She grades 1 math test

our aim is to find how many maths test she grade in 3 hours (120min)

Hence

in 15 mins she does 1 grading

in 120 mins she does x

cross multiply

15x= 120

divide both sides by 15

x= 120/15

x= 8 maths test

Hence in 2 hours, she can grade 8 maths tests

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a rectangular plot of land 8m long and 6m wide.there is a flower bed in the form of a circle of radius 3m.find 1.area of land. 2

mixer [17]

Answer:

1)Area of the Land is 48 square meters

2) Area of flower bed is 28.26 square meter

3)Area of land excluding flower bed is 19.74 square meters

4)cost of planting grass in land excluding flower bed is Rs 118.44

Step-by-step explanation:

Given:

Length of the Rectangular plot = 8m

Width of the rectangular plot = 6m

Radius of the circular flower bed = 3m

1)Area of the Land

Area of the Rectangular plot = Area of the Land

Area of the Land = length x width

Area of the Land = 8*6

Area of the Land = 48 square meters-----------------------------(1)

2. Area of flower bed

Area of flower bed = 4 x area of one quadrant

Area of flower bed = area of the circle

Area of flower bed = $\pi r^2$

where r is the radius

Now , substituting the values

Area of flower bed = $\pi (3)(3)$

Area of flower bed = 28.26 square meters-----------------------------(2)

3)Area of land excluding flower bed

Area of land excluding flower bed = Area of the rectangular land - area of flower beds

Area of land excluding flower bed = 48 - 28.26

Area of land excluding flower bed = 19.74 square meters

4)cost of planting grass in land

Cost of planting grass per square meter = 6

Excluding the flower bed we have 19.74 square meters. to plant grass for this 19. 74 square meters, the cost will be

= $19. 74 \times 6$

= Rs 118.44

6 0

3 years ago

1.Prove each statement. step by step PLEASE!!!

Leya [2.2K]

Answer:

Step-by-step explanation:

a. sin^-1(sin theta) = theta

(1/sin)(sin theta) = theta

the sines cross out theta = theta

b. cos(cos^-1x) = x

cos(x/cos) = x

the cosines cross out x = x

5 0

3 years ago

General solutions of sin(x-90)+cos(x+270)=-1 {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

4 0

3 years ago

Write the ratio 6ft to 4 yd as a fraction in simplest form

vlabodo [156]

6:4 or 3:2 is the answer

5 0

3 years ago

Read 2 more answers

Which expression is equivalent to 8x - 12y + 32

Free_Kalibri [48]

4(2x-3y+8)

Hope its the correct answer

5 0

3 years ago