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Rom4ik [11]
4 years ago
11

Christopher is 20 years younger than Ishaaan. Ishaan and Christopher first met two years ago. Fourteen years ago, Ishaan was 3 t

imes as old as Christopher. How old is Christopher now?
Mathematics
1 answer:
andrey2020 [161]4 years ago
7 0

Answer:

Christopher is 24 years old now

Step-by-step explanation:

I honestly don't know how I got my answer but I know it's correct

You might be interested in
What is the measure of c?<br> A. 20<br> B. 14<br> C. 16<br> D. 22
Harman [31]

Answer:

A

Step-by-step explanation:

The measurements of the segments fall on similar triangles. This means the lengths of each segment can be found using proportions.

Remember, a proportion is an equation which sets equal ratios equal to each other.

\frac{c}{k}=\frac{a}{j}\\\\\frac{c}{32}= \frac{15}{24}

To solve for c, multiply numerators and denominators across the equal sign:

24c = 32(15)

24c = 480

c = 20

5 0
3 years ago
Hard question but would help me out
Alekssandra [29.7K]

Answer:

Critical value f(1)=2.

Minimum at (1,2), function is decreasing for 0 and increasing for x>1.

\left(3,\dfrac{4}{\sqrt{3}}\right) is point of inflection.

When 0<x<3, function is concave upwards and when x>3, , function is concave downwards.

Step-by-step explanation:

1. Find the domain of the function f(x):

\left\{\begin{array}{l}x\ge 0\\x\neq 0\end{array}\right.\Rightarrow x>0.

2. Find the derivative f'(x):

f'(x)=\dfrac{(x+1)'\cdot \sqrt{x}-(x+1)\cdot (\sqrt{x})'}{(\sqrt{x})^2}=\dfrac{\sqrt{x}-\frac{x+1}{2\sqrt{x}}}{x}=\dfrac{2x-x-1}{2x\sqrt{x}}=\dfrac{x-1}{2x^{\frac{3}{2}}}.

This derivative is equal to 0 at x=1 and is not defined at x=0. Since x=0 is not a point from the domain, the crititcal point is only x=1. The critical value is

f(1)=\dfrac{1+1}{\sqrt{1}}=2.

2. For 0 the derivative f'(x)<0, then the function is decreasing. For x>1, the derivative f'(x)>0, then the function is increasing. This means that point x=1 is point of minimum.

3. Find f''(x):

f''(x)=\dfrac{(x-1)'\cdot 2x^{\frac{3}{2}}-(x-1)\cdot (2x^{\frac{3}{2}})'}{(2x^{\frac{3}{2}})^2}=

=\dfrac{2x^{\frac{3}{2}}-2(x-1)\cdot \frac{3}{2}x^{\frac{1}{2}}}{4x^3}=\dfrac{2x^{\frac{3}{2}}-2\cdot\frac{3}{2}x^{\frac{3}{2}}+ 2\cdot\frac{3}{2}x^{\frac{1}{2}}}{4x^3}=

=\dfrac{-x+3}{4x^{\frac{5}{2}}}.

When f''(x)=0, x=3 and f(3)=\dfrac{3+1}{\sqrt{3}}=\dfrac{4}{\sqrt{3}}.

When 0<x<3, f''(x)>0 - function is concave upwards and when x>3, f''(x)>0 - function is concave downwards.

Point \left(3,\dfrac{4}{\sqrt{3}}\right) is point of inflection.

4 0
4 years ago
Please please help me !!
stira [4]

Answer:

your answer is (-12,2)

3 0
3 years ago
Which of the data sets below has a mean of 32? Select all that apply.
Akimi4 [234]

Answer:

B is the correct answer

Step-by-step explanation: When you add 18+54+24=96 divide 96 by 3 and its 32!

5 0
3 years ago
What is the slope of this line graphed below?​
egoroff_w [7]

Answer:

The slope of a linear function. The steepness of a hill is called a slope. The same goes for the steepness of a line. The slope is defined as the ratio of the vertical change between two points, the rise, to the horizontal change between the same two points, the run.

Step-by-step explanation:

make sure you mark me as briliest

5 0
4 years ago
Read 2 more answers
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