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earnstyle [38]
3 years ago
5

Find the slope of the line through the pair of points. (-14, -2) and (7,7)

Mathematics
1 answer:
seraphim [82]3 years ago
5 0

Answer:

3/7

Step-by-step explanation:

Use Y2-Y1/X2-X1

So, 7+2/7+14=9/21

9/21 can be simplified so divide both by 3 which leaves you with 3/7

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JamIah's weight is 4/3 of Devi's weight. What is the ratio of Jamiah's weight to Devi's weight.
melomori [17]

Answer:

Therefore, a ratio of 8/6 is an equivalent ratio of 4/3: in that particular ratio calculation, you should just multiply 4, as well as 3, by 2.

Step-by-step explanation:

I hope this helps! :D

4 0
3 years ago
Read 2 more answers
Five times the sum of a number and 27 is greater than or equal to six times the sum of that
AleksAgata [21]

The complete version of question:

<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26. What is the solution of this problem.</em>

Answer:

5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}

Step-by-step explanation:

As the description of the statement is:

'<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26'.</em>

<em />

As

  • <em>Five times the sum of a number and 27  </em>is written as: 5(x + 27)
  • <em>greater than or equal </em>is written as:  \geq
  • <em>six times the sum of that number and 26'  </em>is written as: 6(x + 26)

so lets combine the whole statement:

<em />5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)<em />

solving

<em />5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)<em />

<em />5x+135\ge \:6x+156<em />

<em />5x+135-135\ge \:6x+156-135<em />

<em />5x\ge \:6x+21<em />

<em />5x-6x\ge \:6x+21-6x<em />

<em />-x\ge \:21<em />

<em />\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}<em />

<em />\left(-x\right)\left(-1\right)\le \:21\left(-1\right)<em />

<em />x\le \:-21<em />

Therefore,

5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}

6 0
3 years ago
In a fruit cocktail, for every 15 ml of orange juice you need 25 ml of apple juice and 10 ml of coconut milk. What proportion of
stepladder [879]

If you sum the amounts of each ingredient, you have a total of

15+25+10 = 50

So, the whole cocktail is 50ml. Of these, 10 are coconut milk. So, the ratio coconut : total is

\dfrac{10}{50} = \dfrac{1}{5}

5 0
3 years ago
Read 2 more answers
In the figure below, if the radius of circle o
Aliun [14]

Answer:

The length of AC is 10 units

Step-by-step explanation:

In the given circle O

∵ AOCB is a rectangle

∵ OB and AC are the diagonals of the rectangle AOCB

∵ Diagonals of the rectangle are equal in lengths

→ That means OB and AC are equal in lengths

∴ OB = AC

∵ O is the center of the circle

∵ B is a point on the circle

∴ OB is a radius of the circle O

∵ The radius of the circle is 10 units

∴ OB = 10 units

∵ OB = AC

∴ AC = 10 units

∴ The length of AC is 10 units

5 0
3 years ago
Evaluate Dx / ^ 9-8x - x2^
Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos &#10;y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
4 0
3 years ago
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