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harkovskaia [24]

2 years ago

8

A tree that is 6 feet tall is growing at a rate of 2 feet every year. A tree that is 12 feet tall is growing at a rate of foot e

ach year. Enter the number of years it will take the two trees to be the same height.

1 answer:

kompoz [17]2 years ago

6 0

Answer:

6 years

Step-by-step explanation:

let x = # of years

6 + 2x = 12 + x

x = 6

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Plz guys help me plz

kari74 [83]

Answer:

option d 11 .......substitute 2 inplace of x

4 0

2 years ago

Factor completely 4x2 − 8x − 60. (4x − 20)(x + 3) 4(x − 5)(x + 3) (4x − 4)(x + 15) 4(x − 1)(x + 15)

katovenus [111]

after complete factorization of $4x^2-8x-60$ the answer is $4(x+3)(x-5)$

Step-by-step explanation:

We need to factor $4x^2-8x-60$

Taking 4 common and factorizing:

$=4(x^2-2x-15)\\=4(x^2+3x-5x-15)\\=4(x(x+3)-5(x+3))\\=4(x+3)(x-5)$

So, after complete factorization of $4x^2-8x-60$ the answer is $4(x+3)(x-5)$

Keywords: factorization, factors

Learn more about factors at:

#learnwithBrainly

3 0

3 years ago

If VW≅VY, WX = s + 69, and XY = 4s, what is the value of s?

Katen [24]

Answer:

s = 23

Step-by-step explanation:

1. VW ≅ VY is given.

2. XV ≅ VX as it is the same side.

3. ∠WVX ≅ ∠YVX as they are both 90°. ∠WVY is 180° as it is a straight line, so both ∠WVX and ∠YVX are exactly half of that.

Given those three things, you can say that ΔWVX ≅ ΔYVX. Because all corresponding parts of congruent triangles are also congruent, you can be sure that WX ≅ XY. Finally, using that information, you can write an equation and solve for s.

$s+69=4s\\s-s+69=4s-s\\69=3s\\\frac{69}{3}=\frac{3s}{3}\\s=23$

7 0

2 years ago

Eric is told by his doctor that he needs to lose 15 pounds. If Eric now weights 285 pounds,then by what percentage should Eric w

tatuchka [14]

Answer:

5.26%

Step-by-step explanation:

15/285=x/100. 15 is the amount of weight he wants to lose. x=5.26%

3 0

2 years ago

Helppp me plsssssssss<br><br>

Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

$\sf \:\:\:\:\:\:\:\:\:\:x_{k}=35$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k}=50$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34$
$\sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42$
$\sf \:\:\:\:\:\:\:\:\:\:h=5$

${\bf \:\: {By\:using\:the\: formula}} \\ \\$

$\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\$

$\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\$

${\large{\frak{\pmb{\underline{Additional\: information }}}}}$

MODE

Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

In a frequency distribution the class having maximum frequency is called the modal class.

${\bf{\underline{Formula\:for\: calculating\:mode:}}} \\$

${\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\$

Where,

$\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.$

$\small \blue{ \bigstar}$ $\sf \: f_{k}=frequency\:of\:the\:modal\:class$

$\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class$

$\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class$

$\small \purple{ \bigstar}$ $\sf \: h= width \:of\:the\:class\:interval$

7 0

3 years ago