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sergejj [24]
3 years ago
11

Show a way to count from 170 to 410 using tens and hundreds. circle at least one benchmark number.

Mathematics
1 answer:
WITCHER [35]3 years ago
6 0
By counting cause you can find your answer duh
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Write a sentence to represent the solution set in the context of the problem.
Nikolay [14]

Answer:

6 sales

Step-by-step explanation:

(I MADE THIS QUESTION I DIDINT KNOW IF YOU NEEDED IT)

Jake wants to but the new computer(it costs $920). Jake has $560 saved in his bank. Jake also gets $60 whenever he makes a sale at his job. How many sale(s) at his job does Jake need to make? <em>Answer </em>-- Jake needs to make 6 sales of $60 to buy the new computer.

MATH -----

$6 x $60 = $360

$360 + $560 = $920

6 0
2 years ago
4. Gloria the grasshopper is working on her hops.
aivan3 [116]

The path that Gloria follows when she jumped is a path of parabola.

The equation of the parabola  that describes the path of her jump is \mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}

The given parameters are:

\mathbf{Height = 20}

\mathbf{Length = 28}

<em>Assume she starts from the origin (0,0)</em>

The midpoint would be:

\mathbf{Mid = \frac 12 \times Length}

\mathbf{Mid = \frac 12 \times 28}

\mathbf{Mid = 14}

So, the vertex of the parabola is:

\mathbf{Vertex = (Mid,Height)}

Express properly as:

\mathbf{(h,k) = (14,20)}

A point on the graph would be:

\mathbf{(x,y) = (28,0)}

The equation of a parabola is calculated using:

\mathbf{y = a(x - h)^2 + k}

Substitute \mathbf{(h,k) = (14,20)} in \mathbf{y = a(x - h)^2 + k}

\mathbf{y = a(x - 14)^2 + 20}

Substitute \mathbf{(x,y) = (28,0)} in \mathbf{y = a(x - 14)^2 + 20}

\mathbf{0 = a(28 - 14)^2 + 20}

\mathbf{0 = a(14)^2 + 20}

Collect like terms

\mathbf{a(14)^2 =- 20}

Solve for a

\mathbf{a =- \frac{20}{14^2}}

\mathbf{a =- \frac{20}{196}}

Simplify

\mathbf{a =- \frac{5}{49}}

Substitute \mathbf{a =- \frac{5}{49}} in \mathbf{y = a(x - 14)^2 + 20}

\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}

Hence, the equation of the parabola  that describes the path of her jump is \mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}

See attachment for the graph

Read more about equations of parabola at:

brainly.com/question/4074088

7 0
3 years ago
Work out the surface area of the solid cuboid. 2.5m 2m 5m The diagram is not drawn to scale.​
kozerog [31]

Answer:

the formula for TSA of a cuboid is (lb+bh+hl)

soo it will be

(2.5x2+2x5+5x2.5)

(5+10+10)

(25)

so 25 cm^2

5 0
2 years ago
A road perpendicular to a highway leads to a farmhouse located d miles away. An automobile traveling on this highway passes thro
pshichka [43]

Answer:

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}

Step-by-step explanation:

A road is perpendicular to a highway leading to a farmhouse d miles away.

An automobile passes through the point of intersection with a constant speed \frac{dx}{dt} = r mph

Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.

Then by Pythagoras theorem,

h² = d² + x²

By taking derivative on both the sides of the equation,

(2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}

(h)\frac{dh}{dt}=(x)\frac{dx}{dt}

(h)\frac{dh}{dt}=rx

\frac{dh}{dt}=\frac{rx}{h}

When automobile is 30 miles past the intersection,

For x = 30

\frac{dh}{dt}=\frac{30r}{h}

Since h=\sqrt{d^{2}+(30)^{2}}

Therefore,

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}

3 0
3 years ago
James has a square piece of paper he wants to cut it into 20 strips of equal with he says this piece of paper is about 48 centim
madam [21]
28 width each strip centimeters
8 0
3 years ago
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