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coldgirl [10]
3 years ago
7

PLSSS HELPP ASAP

Mathematics
1 answer:
dezoksy [38]3 years ago
7 0

Answer:

choose B is the greatest variablity

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What is the linear inequalitie of m=2 over 7
lord [1]

Answer:

Step-by-step explanation:

m=2ov(2.718282)r(7)

Answer:

m=38.055946orv

I think hope this is correct

5 0
3 years ago
What is 15.6 over 100 in simplest form?
marshall27 [118]
\frac{15.6}{100} 
 ← multiply both the numerator and the denominator by 10
\frac{156}{1000} <span>← simplify
</span>\frac{39}{250}

The answer's \frac{39}{250}.
6 0
3 years ago
50 points + brainlest if you answer correctly
Mumz [18]

Answer:

  • <u>5</u><u> </u>is the value which makes the equation true .

Step-by-step explanation:

In this question we have provided an equation that is <u>9</u><u> </u><u>(</u><u> </u><u>3x</u><u> </u><u>-</u><u> </u><u>1</u><u>6</u><u> </u><u>)</u><u> </u><u>+</u><u> </u><u>1</u><u>5</u><u> </u><u>=</u><u> </u><u>6</u><u>x</u><u> </u><u>-</u><u> </u><u>2</u><u>4</u><u> </u>. And we are asked to <u>write </u><u>the </u><u>steps </u><u>to </u><u>solve </u><u>the </u><u>equation </u><u>with </u><u>explanation </u> and <u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>X </u><u>.</u>

<u>Solution</u><u> </u><u>:</u><u> </u><u>-</u>

\longmapsto \quad \: 9(3x - 16) + 15 = 6x - 24

<u>Step </u><u>1</u><u> </u><u>:</u> Solving parenthesis on left side using distributive property which means multiplying 9 with 3x as well as -16 :

\longmapsto \quad \:27x -  \bold{144 }+  \bold{15 }= 6x - 24

<u>Step </u><u>2</u><u> </u><u>:</u> Solving like terms on left side that are -144 and 15 :

\longmapsto \quad \:27x -129 = 6x - 24

<u>Step </u><u>3 </u><u>:</u> Adding 129 on both sides :

\longmapsto \quad \:27x - \cancel{129} -  \cancel{129} = 6x \bold{ - 24 } +  \bold{129}

Now on cancelling -129 with 129 on left side and solving the terms that are -24 and 129 on right side , We get :

\longmapsto \quad \:27x = 6x + 105

<u>Step </u><u>4</u><u> </u><u>:</u> Subtracting with 6x on both sides :

\longmapsto \quad \: \bold{27x} -  \bold{6x} =  \cancel{6x} +105 - \cancel{ 6x}

On calculating further, We get :

\longmapsto \quad \:21x =  105

<u>Step </u><u>5</u><u> </u><u>:</u> Now we are Dividing with 21 on both sides so that we can isolate the variable that is x :

\longmapsto \quad \: \dfrac{ \cancel{21}x}{ \cancel{21}}  = \dfrac{105}{ 21}

Now , by cancelling 21 with 21 on left side , We get :

\longmapsto \quad \:x =  \cancel{\dfrac{105}{21}}

<u>Step </u><u>6</u><u> </u><u>:</u> Now our final step is to simplify the value of x that is 105/21 . We know that 21 × 5 is equal to 105 . So :

\longmapsto \quad \:    \purple{\underline{\boxed{\frak{ x =  5 }}}}

  • <u>Henceforth</u><u> </u><u>,</u><u> </u><u>value </u><u>of </u><u>x </u><u>is </u><u>5</u>

<u>Verifying</u><u> </u><u>:</u><u> </u><u>-</u>

Now we are verifying our answer by substituting value of x in the given equation . So ,

  • 9 ( 3x - 16 ) + 15 = 6x - 24

  • 9 [ 3 ( 5 ) - 16 ] + 15 = 6 ( 5 ) - 24

  • 9 ( 15 - 16 ) + 15 = 30 - 24

  • 9 ( -1 ) + 15 = 6

  • -9 + 15 = 6

  • 6 = 6

  • L.H.S = R.H.S

  • Hence , Verified .

<u>Therefore</u><u>,</u><u> our</u><u> value</u><u> for</u><u> x</u><u> is</u><u> correct</u><u> </u><u>that </u><u>means</u><u> </u><u>it'll</u><u> </u><u>makes</u><u> </u><u>the </u><u>equation</u><u> true</u><u> </u><u>.</u>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
4 0
1 year ago
Which of the following is the midsegment for A ACE?
11111nata11111 [884]

Answer:

A. DB

Step-by-step explanation:

The line BD is the mid segment because it connects the midpoints of line segment AC and CE

5 0
3 years ago
Identify the excluded values of this product. Then rewrite the product in simplest form.
lara [203]

\frac{6 {y}^{2} + 18y - 60 }{3 {y}^{2} - 12y }  \times  \frac{ {y}^{2} - 16 }{ {y}^{2}  + 2y - 8}  \\  \\  \frac{ {6y + 30y - 12y - 60}^{2} }{3y(y - 4)}  \times  \frac{ {(y)}^{2} - ( {4)}^{2}  }{ {y}^{2} + 4y - 2y - 8 }  \\  \\  \frac{6y(y + 5) - 12(y + 5)}{3y(y - 4)}  \times  \frac{(y + 4)(y - 4)}{y(y + 4) - 2(y + 4)}  \\  \\  \frac{(6y - 12)(y + 5)}{3y(y - 4)}  \times  \frac{(y + 4)(y - 4)}{(y + 4)(y - 2)}  \\  \\  \frac{6(y - 2)(y + 5)}{3y(y - 2)}  \\  \\  \frac{2(y + 5)}{y} .

8 0
1 year ago
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