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soldi70 [24.7K]
2 years ago
7

Simplify 3m + 7 – 4m + 6

Mathematics
1 answer:
Ksenya-84 [330]2 years ago
4 0
Subtract or add like terms depending on the sign

3m-4m=-1m
7+6=13
Answer:
-1m+13
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Express 8.5454545454... as a rational number, in the form p/q where p and q are positive integers with no common factors.
zzz [600]
x=8.\overline{54}\\
100=854.\overline{54}\\
100x-x=854.\overline{54}-8.\overline{54}\\
99x=846\\
x=\dfrac{846}{99}=\dfrac{94}{11}


5 0
2 years ago
The length of a rectangular floor is 4 feet longer than its width w. The area of the floor is 525 ft^2. A) Write a quadratic equ
mina [271]

Answer:

x^21x+25x-525-0

x^21x+25x-525-0xx^2 - 3.7_) +5^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21x= -25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21x= -25ensiah193

4 0
2 years ago
If
arlik [135]

Given:

In a right angle triangle θ is an acute angle and \tan\theta =\dfrac{3}{5}.

To find:

The value of \cos \theta.

Solution:

In a right angle triangle,

\tan \theta=\dfrac{Perpendicular}{Base}

We have,

\tan\theta =\dfrac{3}{5}

It means the ratio of perpendicular to base is 3:5. Let 3x be the perpendicular and 5x be the base.

By using Pythagoras theorem,

Hypotenuse=\sqrt{Perpendicular^2+base^2}

Hypotenuse=\sqrt{(3x)^2+(5x)^2}

Hypotenuse=\sqrt{9x^2+25x^2}

Hypotenuse=\sqrt{34x^2}

Hypotenuse=x\sqrt{34}

In a right angle triangle,

\cos \theta=\dfrac{Base}{Hypotenuse}

\cos \theta=\dfrac{5x}{x\sqrt{34}}

\cos \theta=\dfrac{5}{\sqrt{34}}

Therefore, the value of \cos \theta is \dfrac{5}{\sqrt{34}}.

8 0
2 years ago
A shipment of sugar fills 14 containers. If each container holds 2 1/4 tons of sugar, what is the amount of sugar in the entire
SashulF [63]

You would do 14*2 1/4 so that would be 31.5

7 0
3 years ago
Read 2 more answers
webassign If a snowball melts so that its surface area decreases at a rate of 6 cm2/min, find the rate at which the diameter dec
drek231 [11]

Answer:

The answer is \frac{3}{10\pi }  cm/min

Step-by-step explanation:

Assuming the snowball is a perfect sphere, then if A denotes the surface area and D the diameter then:

A=4\pi r^{2} = 4\pi (\frac{D}{2} )^{2} =\pi  D^{2}

Differentiating wrt r we have:

\frac{dA}{dD} =2\pi D

We are told that \frac{dA}{dt}= -6 and we want to find \frac{dD}{dt}

By the chain rule we have:

\frac{dA}{dD}=\frac{dA}{dt}.\frac{dt}{dD}=\frac{\frac{dA}{dt} }{\frac{dD}{dt} }

∴2\pi D=-\frac{6}{\frac{dD}{dt} }

∴\frac{dD}{dt}=-\frac{6}{2\pi D}

When D=10 then

\frac{dD}{dt} =-\frac{6}{10*2\pi } =-\frac{3}{10\pi }

The sign (-) shows that the D is decreasing.

3 0
3 years ago
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