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nika2105 [10]
2 years ago
7

Question b only please

Mathematics
1 answer:
Alisiya [41]2 years ago
8 0

Answer:

total cost= 24.6

Step-by-step explanation:

12 ÷ 20 = 0.6 per straight line

arc 0.6 × 3/2 = 0.9

12 + 0.9 × 14

12 + 12.6 = 24.6

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Write this expression in radical form t^-3/4
ZanzabumX [31]
The first thing you should know are properties of exponents to solve the problem.
 For this case the radical form is given by the writing of the expression in the form of root.
 We have then:
 t^-3/4 =4^root(t^-3)=4^root ((1)/(t^3))
answer  t^-3/4=4^root((1)/(t^3))
5 0
3 years ago
15x ( 7 - 5y ) + 3 ( 5y - x ) - ( 4x + 3 ) ( 5y - 2 )​
Sedbober [7]
15x ( 7 - 5y) + 3 ( 5y - x) - ( 4x + 3) (5y - 2) = -95xy + 110x + 6 is your solution hope it helps :)!
4 0
2 years ago
Read 2 more answers
Solve t/12=4 what is the solution for t?
GaryK [48]

Answer:

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Step-by-step explanation:

4 0
2 years ago
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Determine the resulting polynomial: f ( x ) = 7 x + 9 f(x)=7x+9 g ( x ) = − 5 x 2 − 2 x − 8 g(x)=−5x 2 −2x−8 Find: f ( x ) ⋅ g (
S_A_V [24]

:) Brainliest pls?

Answer:

f(x) * g(x) = -35x^3 - 59x^2 - 74x - 72

Step-by-step explanation:

If f(x) = 7x+9 ang g(x) = -5x^2 - 2x - 8, then

f(x) * g(x) will be:

(7x+9)(-5x^2 - 2x -8)

f(x) * g(x) = -35x^3 - 59x^2 - 74x - 72

7 0
3 years ago
Genetic Inheritance: Calculate probability of two children having the same genotype for achrondoplasia Achondroplasia is a commo
dimaraw [331]

Answer:

The correct answer is - 1/2 or 50% for first and second child to be affected.

Step-by-step explanation:

Achondroplasia is an autosomal dominant disorder. Autosomal dominant disorder refers to the presence of a single copy of the defective gene that is enough to lead to dwarfness.

A cross of achondroplasia (Aa) parent to a person of normal height (aa) result in half of their children will be affected with dwarfism and the other half will be normal.

a  cross between affected or dwarf  and normal parent

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Punnett square:

         a a

A  Aa Aa

a aa aa

Aa- dwarfness

aa- normal height

The probability that both their first child and second child would have achondroplasia is

2/4 =1/2 or 50%.

6 0
3 years ago
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