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Novosadov [1.4K]
3 years ago
11

Han is in training for a marathon. One day, Han ran 8 miles. The next day, he plans to run that same distance plus half as much

again. How far does Han plan to run the next day? *
Mathematics
1 answer:
timofeeve [1]3 years ago
6 0

Answer:

hen is in training of Marathon Vande hand run 8 Miles next day he plans to run the same I bless have much so how far does he planned

Step-by-step explanation:

the hendrun 8 and half miles the next radio Follow Me animal lover

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ipn [44]
1/3 is the correct answer simplified
4 0
2 years ago
Read 2 more answers
What is equivalent to the expression?
Olegator [25]

Answer:

1/16

Step-by-step explanation:

(2^{2})^{-2} \\2^{2*(-2)} \\ 2^{-4} \\ \frac{1}{2^{4} } \\\frac{1}{16}

7 0
3 years ago
Quadrilateral ABCD is inscribed. The measure of ZA = 67º. What is the measure of ZC?
Luda [366]

Answer:

m<C = 113°

Step-by-step explanation:

m<A = 67° (given)

Recall: opposite angles in any inscribed quadrilateral in a circle are equal to 180° or are supplements of one another.

This implies that:

m<A + m<C = 180°

Substitute

67° + m<C = 180°

Subtract 67° from each side

m<C = 180° - 67°

m<C = 113°

7 0
3 years ago
Solve: (3x^2-y)dx + (4y^3-x)dy =0 and find the solution passing through (1,1).
nordsb [41]

Step-by-step explanation:

The given equation is

(3x^{2}-y)dx+(4y^{3}-x)dy=0\\M(x,y)dx+N(x,y)dy=0

As a check for exactness we have

\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial (4y^{3}-x)}{\partial x} =-1\\\\\frac{\partial M}{\partial y}=\frac{\partial (3x^{3}-y)}{\partial y} =-1\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}=-1

Hence the given equation is an exact differential equation and thus the solution is given by

thus the solution is given by

u(x,y)=\int M(x,y)\partial x+\phi (y)\\\\u(x,y)=\int (3x^{2}-y)\partial x+\phi (y,c)\\\\u(x,y)=x^{3}-xy+\phi (y,c)\\\\

Similarly we have

u(x,y)=\int N(x,y)\partial y+\phi (x,c)\\\\u(x,y)=\int (4y^{3}-x)\partial y+\phi (x,c)\\\\u(x,y)=y^{4}-xy+\phi (x,c)\\\\

Comparing both the solutions we infer

\phi (x,c)=x^{3}+c

Hence the solution becomes

u(x,y)=x^{3}+y^{4}-xy=c

given boundary condition is that it passes through (1,1) hence

1^{3}+1^{4}-1=c\\\\\therefore c=1

thus solution is

u(x,y)=x^{3}+y^{4}-xy=1

4 0
3 years ago
Please help me! i linked the image that has the question
ycow [4]

Answer:

Step-by-step explanation:

The area of a trapezoid is the average of the parallel bases times the height.

A=h(a+b)/2 and since we have two identical trapezoids the total are will be

A=h(a+b) in this case

A=15(32+25)=855mm^2

7 0
3 years ago
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