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saveliy_v [14]
1 year ago
7

Help will mark Brainliest!!​

Mathematics
1 answer:
choli [55]1 year ago
3 0

Answer:

A. (h+(h-2) =14)...

I hope this is right! And go on your test! :)

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Yes I love this problem so good
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What is the slope of points (11,13) and (-8,-19)?
borishaifa [10]
The answer is 32/9 , or 1.68.
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Read 2 more answers
Find the scale ratio for the map described below. 1 mm ​(map) equals 500 m ​(actual) The scale ratio is 1 to nothing.
lana66690 [7]

Answer:

1:500000

Step-by-step explanation:

1 mm ​(map) equals 500 m ​(actual) .

Let's convert 500 m to mm.

1m = 1000mm

500m = 500000 mm

So 1mm to 500000mm on a scale is

1:500000

So it's all about converting the metre to million metre then doing the ratio.

In this case we are not to divide anything because it's already in 1.

So it's 1mm on paper then 500000mm on actual.

Thank you

4 0
3 years ago
Simon put $15000 in an account with a simple interest rate of 4.5%. How much interest will be earned after 8 years?
Leto [7]

~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$15000\\ r=rate\to 4.5\%\to \frac{4.5}{100}\dotfill &0.045\\ t=years\dotfill &8 \end{cases} \\\\\\ I = (15000)(0.045)(8)\implies I=5400

4 0
2 years ago
Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3> (1 point)
sesenic [268]

Angle between u = -5i-4j , v=-4i-3j is x =0° .

<u>Step-by-step explanation:</u>

We have , two vectors u = <-5, -4>, v = <-4, -3>  or , u = -5i-4j , v=-4i-3j

We need to find angle between these two vectors . Let's find out:

We know that dot product of two vectors is defined as :

u.v =|u|(|v|)cosx , where x is angle between u & v !

⇒ u.v =|u|(|v|)cosx

⇒ cosx =\frac{u.v}{|u|(|v|)}

Now , u.v = (-5i-4j)(-4i-3j)

⇒ u.v = (-5i-4j)(-4i)-(-5i-4j)(3j)

⇒ u.v = 20+12            { i(j) = j(i) =0  }

⇒ u.v = 32

Now , Modulus of any vector  r = xi+yj is |r| = \sqrt{x^{2}+y^{2}} So ,

|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5

Putting all these values in equation cosx =\frac{u.v}{|u|(|v|)} we get:

⇒ cosx =\frac{32}{5(\sqrt{41})}

⇒ cos^{-1}(cosx) =cos^{-1}(\frac{32}{5(6.4)})

⇒ x =cos^{-1}(1)                 { cos0 = 1  }

⇒ x =0°

Therefore , Angle between u = -5i-4j , v=-4i-3j is x =0° .

8 0
3 years ago
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