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Sedbober [7]
4 years ago
12

Please Help!! Urgent ( Zoom in to see it better)

Mathematics
1 answer:
ruslelena [56]4 years ago
8 0

1. You are trying to isolate "F" in the equation

C=\frac{5}{9}(F-32)      Multiply 9/5 on both sides

(\frac{9}{5})C=(\frac{9}{5}) \frac{5}{9} (F-32)

\frac{9}{5}C=F-32    Add 32 on both sides

\frac{9}{5}C+32=F-32+32

\frac{9}{5}C+32=F

THIS IS WRONG


2. You are trying to isolate "y" in the equation

m=\frac{x+y+z}{3}   Multiply 3 on both sides

3m = x + y + z        Subtract x and z on both sides

3m - x - z = y

THIS IS CORRECT


3. Isolate "r"

s=\frac{r}{r-1}   Multiply (r - 1) on both sides

s(r - 1) = r    Distribute s into (r - 1)

sr - s = r     Subtract sr on both sides

-s = r - sr    Factor out r from (r - sr)

-s = r(1 - s)    Divide (1 - s) on both sides

\frac{-s}{1-s}=r

THIS IS WRONG


4. Isolate "b"

A=\frac{1}{2}(a+b)    Multiply 2 on both sides

2A = a + b     Subtract a on both sides

2A - a = b

THIS IS CORRECT


5. Isolate "y"

m=\frac{x+y}{2}    Multiply 2 on both sides

2m = x + y    Subtract x on both sides

2m - x = y

THIS IS WRONG


The 2nd and 4th one is right

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Please help with this math question.​
V125BC [204]

The coordinate grid of the ordered pair is gotten from the combination of the y and X axis of the graph.

<h3>Finding of the ordered pair</h3>

Using the coordinates, an ordered pair for the given point is:

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  • H= (10,8) because the 10 corresponds to the point in y axis and 8 on the X axis.

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7 0
2 years ago
Find two integers whose product is 138 such that one of the integers is seven less than five times the other integer.
Oksana_A [137]

Answer: the integers are 6 and 23

Step-by-step explanation:

Let x represent one of the integers.

Let y represent the other integer.

The product of the integers is 138. This means that

xy = 138 - - - - - - - - - -1

One of the integers is seven less than five times the other integer. This means that

x = 5y - 7

Substituting x = 5y - 7 into equation 1, it becomes

y(5y - 7) = 138

5y² - 7y = 138

5y² - 7y - 138 = 0

5y² + 23y - 30y - 138 = 0

y(5y + 23) - 6(5y + 23)

y - 6 = 0 or 5y + 23 = 0

y = 6 or y = - 23/5

The only possible value of y is 6

x = 138/y = 138/6

x = 23

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3 years ago
a cuboid with a volume of 924 cm3 has dimensions 4cm (x+1)cm and (x+11)cm. show clearly that x^2 +12x-220=0. show the equation b
nikdorinn [45]

Answer:

Part 1)

The equation by factorization is

x^{2}+12x-230=(x-(-6+\sqrt{266}))(x-(-6-\sqrt{266}))

Both values of x are

x1=-6+\sqrt{266}

x2=-6-\sqrt{266}

Part 2) The dimensions of the cuboid are

4\ cm, (-5+\sqrt{266})\ cm and (5+\sqrt{266})\ cm

Step-by-step explanation:

step 1

we know that

The volume of the cuboid is equal to

V=LWH

substitute the given values

V=4(x+1)(x+11)

V=924\ cm^{3}

so

4(x+1)(x+11)=924\\4(x^{2}+11x+x+11)=924\\ 4(x^{2}+12x+11)=924\\x^{2}+12x+11=231\\x^{2}+12x-230=0

Complete the square

x^{2}+12x-230=0

x^{2}+12x=230

x^{2}+12x+36=230+36

(x+6)^{2}=266

square root both sides

x+6=(+/-)\sqrt{266}

x=-6(+/-)\sqrt{266}

so

x1=-6+\sqrt{266}

x2=-6-\sqrt{266}

therefore

The equation by factorization is

x^{2}+12x-230=(x-(-6+\sqrt{266}))(x-(-6-\sqrt{266}))

step 2

Find the dimensions of the cuboid

The dimensions are

4\ cm

(x+1)\ cm -----> (-6+\sqrt{266}+1)=(-5+\sqrt{266})\ cm

(x+11)\ cm ----> (-6+\sqrt{266}+11)=(5+\sqrt{266})\ cm

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