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ipn [44]
3 years ago
8

What scale is used on the graph below

Mathematics
1 answer:
kondor19780726 [428]3 years ago
6 0

\orange{\underline{{\bold{\mathbb{\blue{  \huge{QUESTION}}}}}}}

What scale is used on the graph

\huge \bf {\red A \pink{N}\purple{S} \blue{W} \orange{ER}}

b). 2 cm to 2 units on both axes

As we can se there is gap of 2 boxes (side length= 1cm of one box so 2 boxes side length= 2cm) between 2 units so we can conclude that scale is equal to 2 cm to 2 units on both axes

\bold \red{DEFINATION  \:  \: OF  \:  \: SCALE }

IT IS THE RELATIONSHIP BETWEEN THE UNIT USED AND ACTUAL VALUE ON THE GRAPH

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What is the length of the hypotenuse
nika2105 [10]

<u>Given</u>:

Given that the length of the two legs of the right triangle are 28 m and 45 m.

We need to determine the length of the hypotenuse.

<u>Length of the hypotenuse:</u>

The length of the hypotenuse can be determined using the Pythagorean theorem.

Thus, we have;

hyp^2=28^2+45^2

Simplifying, we get;

hyp^2=784+2025

hyp^2=2809

Taking square root on both sides, we get;

hyp=\sqrt{2809}

hyp=53 \ m

Thus, the length of the hypotenuse is 53 m

Hence, Option C is the correct answer.

5 0
3 years ago
Read 2 more answers
The x-intercepts and vertex of the quadratic relation y = –(x + 3)(x + 5) are:
baherus [9]

Answer:

d) x-intercepts are –3 and –5; vertex is (–4, 1)

Step-by-step explanation:

parabola are the vertex findings -4

we find these using formula m +n  / 2

m+ n = -3 -5 / 2 = -4

m+n / 2 = -8/2 = -4

xv = -4

plug in -4 to find yv

(x+5)

(-4 + 5)

yv = 1

5 0
2 years ago
Use integration by parts to find the integrals in Exercise.<br> ∫^3_0 3-x/3e^x dx.
Viefleur [7K]

Answer:

8.733046.

Step-by-step explanation:

We have been given a definite integral \int _0^3\:3-\frac{x}{3e^x}dx. We are asked to find the value of the given integral using integration by parts.

Using sum rule of integrals, we will get:

\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx

We will use Integration by parts formula to solve our given problem.

\int\ vdv=uv-\int\ vdu

Let u=x and v'=\frac{1}{e^x}.

Now, we need to find du and v using these values as shown below:

\frac{du}{dx}=\frac{d}{dx}(x)

\frac{du}{dx}=1

du=1dx

du=dx

v'=\frac{1}{e^x}

v=-\frac{1}{e^x}

Substituting our given values in integration by parts formula, we will get:

\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(x*(-\frac{1}{e^x})-\int _0^3(-\frac{1}{e^x})dx)

\frac{1}{3}\int _0^3\frac{x}{e^x}dx=\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))

\int _0^3\:3dx-\int _0^3\frac{x}{3e^x}dx=3x-\frac{1}{3}(-\frac{x}{e^x}- (\frac{1}{e^x}))

Compute the boundaries:

3(3)-\frac{1}{3}(-\frac{3}{e^3}- (\frac{1}{e^3}))=9+\frac{4}{3e^3}=9.06638

3(0)-\frac{1}{3}(-\frac{0}{e^0}- (\frac{1}{e^0}))=0-(-\frac{1}{3})=\frac{1}{3}

9.06638-\frac{1}{3}=8.733046

Therefore, the value of the given integral would be 8.733046.

6 0
3 years ago
Find the exact length of the third side.
frez [133]

Step-by-step explanation:

Hello there!

Use the Pythagoream theorem:

a^{2} +b^{2} =c^{2} \\4^2+2^2=c^2\\16+4=20\\x=\sqrt{20} \\x=4.472135

:)

7 0
3 years ago
Read 2 more answers
Felipe has a job transporting soft drinks by truck. his truck is filled with cans that weigh 14 ounces each and bottles that wei
Ray Of Light [21]
Hello,

Let x be the number of 14-ounce cans

800-x be the number of bottles

Total weight=14*x+70*(800-x)=14x+56000-70x=56000-56x=56(1000-x)
5 0
3 years ago
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