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3 years ago

NEED HELP PLEASE HELP AS FAST AS YOU CAN!!!!! MATH:(

1 answer:

3 0

In any equation with variables, you want to end with a variable isolated on one side with no coefficients(like x=4, for instance).

So, with that in mind, let's solve number 2.

5(x-3) - 2(2x+4) = -2(x-2)
Let's distribute the coefficients.
(5x-15)-(4x+8) = (-2x+4)
Simplify the left side first (by subtracting the two)
5x-4x = x
-15 - (+)8 = -23
So, left side is
x-23 = -2x+4
We're close now, just bring the x to the left side and the other values to the right
x-23 = -2x+4
+23 +23
x=-2x+27
+2x +2x
3x=27
divide by 3
x=9

3.
3(4x-5)-2(2x-2)=4(3x-2)+5
Same idea here; distribute the coefficients, and simplify.
(12x-15)-(4x-4)=(12x-8)+5
12x-4x=8x
-15-(-)4=-11
-8+5=-3
8x-11=12x-3
bring x to left, other numbers to right
-12x -12x
+11 +11
-4x = 8
divide by -4
x=-2

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Tavon has a gift card for $70 that loses $3 for each 30-day period it is not used. He has another gift card for $ 60 that loses

vladimir2022 [97]

Hello,

Your answer would be: 55

My work:

For this problem all you have to do is keeping subtracting 3 until you get a matching number in which I got 55.

Hope this helps!

Brainliest!?!

Have a great day:)

~Rendorforestmusic

5 0

3 years ago

The following tables each show the inventory of a warehouse and the number of orders for four different products which have been

snow_lady [41]

Orders: x
Inventory: y

1) First table
x2-x1=6-3→x2-x1=3
y2-y1=1920-1960→y2-y1=-40

x3-x2=9-6→x3-x2=3=x2-x1
y3-y2=1900-1920→y3-y2=-20 different to y2-y1=-40. The table does not represent a linear relationship.

2) Second table
x2-x1=7-5→x2-x1=2
y2-y1=1860-1900→y2-y1=-40

x3-x2=9-7→x3-x2=2=x2-x1
y3-y2=1820-1860→y3-y2=-40=y2-y1

x4-x3=11-9→x4-x3=2=x3-x2
y4-y3=1780-1820→y4-y3=-40=y3-y2

x5-x4=13-11→x5-x4=2=x4-x3
y5-y4=1740-1780→y5-y4=-40=y4-y3

The table represents a linear relationship.

3) Third table
x2-x1=2-1→x2-x1=1
y2-y1=1000-2000→y2-y1=-1000

x3-x2=3-2→x3-x2=1=x2-x1
y3-y2=500-1000→y3-y2=-500 different to y2-y1=-1000. The table does not represent a linear relationship.

4) Fourth table
x2-x1=6-4→x2-x1=2
y2-y1=1640-1840→y2-y1=-200

x3-x2=8-6→x3-x2=2=x2-x1
y3-y2=1360-1640→y3-y2=-280 different to y2-y1=-200. The table does not represent a linear relationship.

Answer: The second table best represents a linear relationship.

8 0

4 years ago

Read 2 more answers

Last question was inncorrectly typed:1.A sample of students who has taken a calculus test has a mean =78.2, mode =67, median 67.

vampirchik [111]

Answer:

(A) The data set of the sample of students who has taken a calculus test is right skewed.

(B) The average of female students in the class is 66.

Step-by-step explanation:

(A)

For a left skewed distribution the Mean < Median < Mode.

For a right skewed distribution the Mean > Median > Mode.

For a symmetric distribution the Mean = Median = Mode.

Given: Mean = 78.2, Median = 67 and Mode = 67

In this case the mean of the data is more than the median and mode.

$Mean = 78.2>Median = Mode = 67$

Thus, the data set of the sample of students who has taken a calculus test is right skewed.

(B)

Total number of student ( $n$ ) = 35

Combined average ( $\mu_{c}$ ) = 70

Number of male student ( $n_{M}$ ) = 20

Average of male students ( $\mu_{M}$ ) = 73

Number of female student ( $n_{F}$ ) = 15

Average of female students = $\mu_{F}$

The formula to compute the combined average is:

$\mu_{c}=\frac{n_{M}\mu_{M}+n_{F}\mu_{F}}{n_{M}+n_{F}}$

Compute the value of $\mu_{F}$ as follows:

$\mu_{c}=\frac{n_{M}\mu_{M}+n_{F}\mu_{F}}{n_{M}+n_{F}}\\70=\frac{(20\times73)+(15\times\mu_{F})}{20+15}\\ 70\times35=1460+(15\times\mu_{F})\\\mu_{F}=\frac{2450-1460}{15} \\=66$