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Jlenok [28]
3 years ago
14

I don't get this please help me! it's only 8th grade math. thanks!(:

Mathematics
1 answer:
solmaris [256]3 years ago
8 0
The time × 1.5=the distance
so 5×1.5=7.5
10×1.5=15
20×1.5=30
im a 7th grader doing 8th grade math.. just think
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An anthropologist studies a woman's femur that was uncovered in Madagascar. To estimate the woman's height, he uses the equation
Norma-Jean [14]

Answer:

The inequality best represents the length of the femur that would suggest the woman had a height greater than 160 cm is:

f > 40

Step-by-step explanation:

Woman's height (cm): h = 60 + 2.5 f

Length of the femur (cm): f


Height greater than 160 cm:

h > 160

Replacing h by 60+2.5f in the inequality above:

60+2.5f > 160

Solving for f: Subtracting 60 both sides of the inequality:

60+2.5f-60 > 160-60

Subtracting

2.5f > 100

Dividing both sides of the inequality by 2.5:

2.5f/2.5 > 100/2.5

f > 40

Answer: The inequality best represents the length of the femur that would suggest the woman had a height greater than 160 cm is:

f > 40


5 0
3 years ago
Identify the oblique asymptote of f(x) = quantity 3 x squared plus 2x minus 5 over quantity x minus 4.
NemiM [27]
F(x)=(3x^2+2x-5)/(x-4)
f(x)=(3x+5)(x-1)/(x-4)
x-4|(3x^2+2x-5)
(x-4)3x
3x^2+2x-5-(3x^2-12x)
(x-4)3x+14
14x-5-(14x-56)
51
Oblique=3x+14
8 0
3 years ago
Read 2 more answers
Solve the equations 1. 16x^2-1=0 2. 36z^2 96z 15=0
Allisa [31]
16x^2-1=0 2 

16x^2-1=0 2:x 


16x^2=1
 
   |Add \ 1 \ to \ both \ sides|

x^2= \dfrac{1}{16}   |Divide \ both \ sides \ by \ 16|<span>
</span>
x^2= \sqrt{\dfrac{ \sqrt{1}}{ \sqrt{16}}}  |Take \ the \ square \ root \ of \ both \ sides|

x= \dfrac{1}{4}


5 0
3 years ago
CAN SOMEONE HELP ME ANSWER THIS ?
kumpel [21]
Use gauthmath it can help u it’s any math work instead of waiting on here 3EMB3R use that code to get more tickets
7 0
3 years ago
WILL GIVE BRAINIEST TO WHOEVER GETS IT RIGHT. PLSHELP THX.
ASHA 777 [7]

Answer:

we get rectangular coordinates,

x= \frac{3}{2}

y =  \frac{-3√3}{2}

Step-by-step explanation:

We have given,

Polar coordinates ,P = (r , θ) = (3 , \frac{-pi}{3} )

We nee to find rectangular coordinates:

Since , x =rcosθ  and y =  rsinθ

i.e x = 3cos(frac{-pi}{3}[/tex]) = 3×\frac{1}{2} = \frac{3}{2}

And y = 3sin(frac{-pi}{3}[/tex]) = 3×\frac{-√3}{2} = \frac{-3√3}{2}

Hence we get rectangular coordinates,

x= \frac{3}{2}

y =  \frac{-3√3}{2}

7 0
3 years ago
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