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3 years ago

Please HELP!!!!!

Mathematics

1 answer:

Stels [109]3 years ago

3 0

Can you explain it more clear because my english is not very good, I want to easy to help you

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1. The quadratic parent function f(x) = x2 was transformed to

AlekseyPX

Answer:

Step-by-step explanation:

Given f(x) then f(x + a) is a horizontal translation of f(x)

• If a > 0 then a shift left of a units

• If a < 0 then a shift right of a units

The graph of g(x) is the graph of f(x) shifted 6 units left , then

g(x) = f(x + 6) → C

7 0

3 years ago

X² + 5x - 24, factor

kobusy [5.1K]

Answer:

${x}^{2} - 8x + 3x - 24 \\ - x(x - 8)3(x - 8) \\ (3 - x)(x - 8)$

6 0

3 years ago

Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

pychu [463]

Answer:

A=152

K= -Ln(0.5)/14

Step-by-step explanation:

You can obtain two equations with the given information:

T(14 minutes) = 114◦C

T(28 minutes)=152◦C

Therefore, you have to replace t=14, T=114 and t=28, T=152 in the given equation:

$114=190-Ae^{-14k} (I) \\152=190-Ae^{-28k}(II)$

Applying the following property of exponentials numbers in (II):

$e^{a}.e^{b}=e^{a+b}$

Therefore $e^{-28k}$ can be written as $e^{-14k}.e^{-14k}$

$152=190-Ae^{-14k}.e^{14k}$

Replacing (I) in the previous equation:

$152=190-76e^{-14k}$

Solving for k:

Subtracting 190 both sides, dividing by -76:

$0.5=e^{-14k}$

Applying the base e logarithm both sides:

Ln(0.5)= -14k

Dividing by -14:

k= -Ln(0.5)/14

Replacing k in (I) and solving for A:

$Ae^{-14(-Ln(0.5)/14)}=76\\Ae^{Ln(0.5)} =76\\A(0.5)=76$

Dividing by 0.5

A=152

7 0

3 years ago

Ben claims that 12 is a factor of 24. how can you check whether he is correct

AlekseyPX

Take 12 from 24 and if the the answer is 12 then it's a factor.

7 0

3 years ago

Match its group with its percentage or mean Using

krok68 [10]

(1)

Mean length of all fish in the sample - $\boxed{\sf{\red{6.2}}}$

Mean = (Sum of observations)/(Number of observations)

= (12 + 5 + 3 + 5 + 8 + 2 + 10 + 9 + 4 + 4)/(10)

= 62/10

= 6.2

(2)

Mean length of adult fish in the sample - $\boxed{\sf{\red{8.75}}}$

Mean = (Sum of observations)/(Number of observations)

= (12 + 5 + 8 + 10)/(4)

= 35/4

= 8.75

(3)

Mean length of juvenile fish in the sample - $\boxed{\sf{\red{4.5}}}$

Mean = (Sum of observations)/(Number of observations)

= (5 + 3 + 2 + 9 + 4 + 4)/(6)

= 27/6

= 4.5

(4)

Percentage of sample that were adult fish - $\boxed{\sf{\red{40}}}$

Percentage = (No. of adult fishes)/(Total no. of fishes) × 100

% = (4/10) × 100

% = 40

(5)

Percentage of sample that were juvenile fish - $\boxed{\sf{\red{60}}}$

Percentage = (No. of juvenile fishes)/(Total no. of fishes) × 100

% = (6/10) × 100

% = 60

(6)

Percentage of sample that were juveniles over 8 inches long - $\boxed{\sf{\red{10}}}$

Percentage = (No. of juveniles over 8 inches)/(Total no. of fishes) × 100

% = (1/10) × 100

% = 10

7 0

3 years ago