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3 years ago

13

Mary compares the heights of two trees. Their heights are shown. Write and solve an equation to find the height h of the taller

tree.

1 answer:

Mariana [72]3 years ago

3 0

45+12= the height of the taller tree

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Put thes fallowing numbers in order from largest to smallest -3, -3.5, 6, -6, 0, 1

Verdich [7]

6, 1, 0, -3, -3.5, -6

5 0

3 years ago

Read 2 more answers

Does anyone know what to fill in?

riadik2000 [5.3K]

Answer:

$36.69

Step-by-step explanation:

Okay, so let's first write the equation:

2.59*10 + 3d = 135.97

Now, let's work on isolating d by first simplifying the equation:

25.9 + 3d = 135.97

25.9 + 3d - 25.9 = 135.97 - 25.9

3d = 110.07

3d/3 = 110.07/3

d = 36.69

Okay, now let's check:

2.59*10 + 3*36.69 = 135.97

25.9 + 110.07 = 135.97

135.97 = 135.97

Okay, so so it costs $36.69 per pair of shoe.

4 0

3 years ago

ASAP! GIVING BRAINLIEST! Please read the question THEN answer CORRECTLY! NO guessing. I say no guessing because people usually g

inna [77]

Answer:

D. It would be less steep

Step-by-step explanation:

The first graph moves at a rate of 5/1 which is a greater fraction than 3/4

The second graph is shallow due to the close points in x and y that are able to be conducted
The first Graph rapidly increases at a way higher rate making it VERY steep
While both are linear the second strays away in terms of plot lines

7 0

3 years ago

Use the Fundamental Theorem of Calculus to find the area of the region between the graph of the function x5 + 8x4 + 2x2 + 5x + 1

belka [17]

Answer:

The area of the region between the graph of the given function and the x-axis = 25,351 units²

Step-by-step explanation:

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15

If 'f' is a continuous on [a ,b] then the function

$F(x) = \int\limits^a_b {f(x)} \, dx$

By using integration formula

$\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c$

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

$\int\limits^6_^-6} (x^{5} + 8 x^{4} + 2 x^{2} + 5 x + 15) )dx$

<em>On integration , we get</em>

= $(\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}$

$F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)$

= $(\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))$

After simplification and cancellation we get

= $\frac{2 X 8 X (6)^{5} }{5} + \frac{2 X 2 X (6)^3}{3} + 2 X 15 X 6$

on calculation , we get

= $\frac{124,416}{5} + \frac{864}{3} + 180$

On L.C.M 15

= $\frac{124,416 X 3 + 864 X 5 + 180 X 15}{15}$

= 25 351.2 units²

<u><em>Conclusion</em></u>:-

<em>The area of the region between the graph of the given function and the x-axis = 25,351 units²</em>

6 0

3 years ago

A coat at a store is originally priced at $54.99. The store marks the price down to $32.99. To the nearest percent, what is the

Dmitry_Shevchenko [17]

Answer:

66.6%

Step-by-step explanation:

Step one

given

A coat at a store is originally priced at $54.99

The store marks down the price down to $32.99

Required

the percent decrease

Step two:

the percent decrease= change in price/ old price*100

the percent decrease= 54.99-32.99/54.99*100

the percent decrease= 22/ 32.99*100

the percent decrease= 0.666*100

the percent decrease= 66.6%

6 0

3 years ago