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Alex17521 [72]
2 years ago
13

Need help if you can, help please help

Mathematics
2 answers:
Nana76 [90]2 years ago
5 0

Answer:

4.4 km/h

Step-by-step explanation:

From the graph you can see she was at the shop after 30 minutes. If you travel 2.2 km in 30 minutes, your speed is 2.2 / 0.5 = 4.4 km/h

So the trick is to express the 30 minutes as 1/2 hour.

kramer2 years ago
3 0

Answer:

4.4 km/h

Step-by-step explanation:

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earnstyle [38]

Answer: -2

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Is 3/10 greater than or less than 1/2
V125BC [204]
To compare two fraction we have to make common denominator, then we can compare numerators.
Lets do this
\frac{1}{2}= \frac{1*5}{2*5}= \frac{5}{10}
Now we can compare numeratos.
3<5, so 1/2 is greater than 3/10.
8 0
3 years ago
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What is the value of the expression 2 x 33 + 17 + 5 x 8
snow_lady [41]

Answer:

Correct answer of this question is 123

5 0
2 years ago
Find the measures of the three angles, in radians, of the triangle with the given vertices: d(1,1,1), e(1,−5,2), and f(−2,2,7).
Oduvanchick [21]

Consider triangle DEF with vertices D(1,1,1), E(1,-5,2) and F(-2,2,7).

1. Find

\overrightarrow{DE}=(1-1,-5-1,2-1)=(0,-6,1),\\ \\\overrightarrow{DF}=(-2-1,2-1,7-1)=(-3,1,6).

Then

\cos \angle D=\dfrac{0\cdot (-3)+(-6)\cdot 1+1\cdot 6}{\sqrt{0^2+(-6)^2+1^2}\cdot \sqrt{(-3)^2+1^2+6^2}}=\dfrac{0}{\sqrt{37} \cdot \sqrt{46} }=0.

2. Find

\overrightarrow{ED}=(1-1,1-(-5),1-2)=(0,6,-1),\\ \\\overrightarrow{EF}=(-2-1,2-(-5),7-2)=(-3,7,5).

Then

\cos \angle E=\dfrac{0\cdot (-2)+6\cdot 7+(-1)\cdot 5}{\sqrt{0^2+6^2+(-1)^2}\cdot \sqrt{(-3)^2+7^2+5^2}}=\dfrac{37}{\sqrt{37} \cdot \sqrt{83} }=\sqrt{\dfrac{37}{83}}.

3. Find

\overrightarrow{FE}=(1-(-2),-5-2,2-7)=(3,-7,-5),\\ \\\overrightarrow{FD}=(1-(-2),1-2,1-7)=(3,-1,-6).

Then

\cos \angle F=\dfrac{3\cdot 3+(-7)\cdot (-1)+(-5)\cdot (-6)}{\sqrt{3^2+(-7)^2+(-5)^2}\cdot \sqrt{3^2+(-1)^2+(-6)^2}}=\dfrac{46}{\sqrt{83} \cdot \sqrt{46} }=\sqrt{\dfrac{46}{83}}.

4.

\angle E=\arccos0=\dfrac{\pi}{2},\\ \\&#10;\angle D=\arccos\letf(\sqrt{\dfrac{37}{83}}\right)\approx 0.27\pi,\\ \\&#10;\angle F=\arccos\letf(\sqrt{\dfrac{46}{83}}\right)\approx 0.23\pi.

3 0
2 years ago
25. angle EFG and angle GFH are a linear pair,
vivado [14]

Answer:

m<EFG = 69^{o}, and m<GFH = 111^{o}

Step-by-step explanation:

Linear pair angles are two supplementary angles.

Thus,

m<EFG + m<GFH = 180^{o}

2n + 21 + 4n + 15 = 180^{o}

collecting like terms, we have:

6n + 36 = 180^{o}

6n = 180^{o} - 36

6n = 144^{o}

divide both both sides by 6,

n = 24^{o}

Therefore,

m<EFG = 2n + 21

            = 2 x 24^{o} + 21

            = 48 + 21

            = 69^{o}

m<GFH = 4n + 15

             = 4 x 24^{o} + 15

            = 96 + 15

            = 111^{o}

Thus m<EFG = 69^{o}, and m<GFH = 111^{o}

7 0
3 years ago
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