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mart [117]
3 years ago
13

-(10)^-2 simplify the expression

Mathematics
2 answers:
STatiana [176]3 years ago
8 0

your answer is -1/100 (a fraction)

kiruha [24]3 years ago
6 0

The final answer is -1/100.

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AB = 10 in; AD = 6 in <br><br> A= <br><br> 30sqrt3 in2<br> 30 in2<br> 60 in2
Ratling [72]

Answer: 30

Step-by-step explanation:

If we were to draw diagonal BD, then we would create two congruent triangles, and we could find the area of each using the formula A=\frac{1}{2} ab \sin C.

However, since there are two triangles, we can double this, giving us A=ab \sin C.

So, the area is A=(10)(6)(\sin 30^{\circ})=\boxed{30}

8 0
1 year ago
He charges $15 per hour to mow lawns and $20 per hour for gardening.
Reil [10]

Answer:

Step-by-step explanation:

he can make 300 dollars per week by doing either or. so if you're asking whether the statement is correct, it is

3 0
2 years ago
Find the domain and range of the exponential function h(x) = –343x.
Amanda [17]

The function is supposed to be;

h(x) = -343^(x)

Answer:

The domain will be a set of real numbers while the range will be y ≤ 0 and on the interval (-∞, 1)

As x is increasing, h is decreasing and as x is decreasing, h is increasing

Step-by-step explanation:

We are given the function as;

h(x) = -343^(x)

The formula is;

y = a^(x) since it's symmetrical to the x-axis

However in this case;

y = -a^(x)

Now, the domain is y and the range is a set of values of x.

I've attached a graph of this function drawn on desmos.

From the graph we can see that The domain will be a set of real numbers while the range will be on the interval (-∞, 1)

For a value of x = 0,we have;

h(0) = -343^(0)

h(0) = -343

When we increase the value of x to 3,we have;

h(3) = -343^(3)

h(3) = -40353607

When we decrease the value of x to -3, we have;

h(-3) = -343^(-3)

h(-3) = 0.00000002478

Thus, we can conclude that;

As x is increasing, h is decreasing and as x is decreasing, h is increasing

4 0
2 years ago
Write an equation in standard form of the hyperbola described.
marishachu [46]

Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.

\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill

\begin{cases} h=0\\ k=0\\ a=2\\ c=4 \end{cases}\implies \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{b^2} \\\\\\ c^2=a^2+b^2\implies 4^2=2^2+b^2\implies 16=4+b^2\implies \underline{12=b^2} \\\\\\ \cfrac{(x-0)^2}{2^2}-\cfrac{(y-0)^2}{12}\implies \boxed{\cfrac{x^2}{4}-\cfrac{y^2}{12}}

5 0
2 years ago
NEED ANSWER FAST PLASE FAST
Dovator [93]

Answer:

D. no solution

Step-by-step explanation:

h-z=3(1)

h=3+z(minus z from both side)

put in the second equation

-3h,+3z=6(2)

-3(3+z)+3z=6

-9-3z+3z=6

-9=6( not true)

so the equation doesn't have a solution.

3 0
2 years ago
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