All categories

3 years ago

Help!!!!

Mathematics

1 answer:

Feliz [49]3 years ago

6 0

Answer:

Step-by-step explanation:

<h3>Exponential Form :</h3><h3>(a^m*a^n ) = (a^m+n )</h3>

<h3>(only 7^-4 the term that can add to 7^-2 gives </h3><h3>= 7^-6 )</h3>

You might be interested in

Is it ever possible for the domain and range in a function to have different numbers of entries (for example, 3 domain entries t

wariber [46]

No , & More Than Likely If It Is To Happen It's Because The Domain Entry Repeated A Number Mulitiple Times

5 0

3 years ago

laurie will draw a scale model of the garden she wants to plant. Her scale will be 1cm = 2.5 ft. what will be the actual dimensi

g100num [7]

Answer:

Step-by-step explanation:

Because i said.......

4 0

3 years ago

Read 2 more answers

Scientists modeled the intensity of the sun, I, as a function of the number of hours since 6:00 a.m., h, using the

MAVERICK [17]

The functions are illustrations of composite functions.

<em>The soil temperature at 2:00pm is 67</em>

The given parameters are:

$\mathbf{I(h) =\frac{12h - h^2}{36}}$ ---- the function for sun intensity

$\mathbf{T(I) =\sqrt{5000I}}$ -- the function for temperature

At 2:00pm, the value of h (number of hours) is:

$\mathbf{h = 2:00pm - 6:00am}$

$\mathbf{h = 8}$

Substitute 8 for h in $\mathbf{I(h) =\frac{12h - h^2}{36}}$ , to calculate the sun intensity

$\mathbf{I(8) =\frac{12 \times 8 - 8^2}{36}}$

$\mathbf{I(8) =\frac{32}{36}}$

$\mathbf{I(8) =\frac{8}{9}}$

Substitute 8/9 for I in $\mathbf{T(I) =\sqrt{5000I}}$ , to calculate the temperature of the soil

$\mathbf{T(8/9) =\sqrt{5000 \times 8/9}}$

$\mathbf{T(8/9) =\sqrt{4444.44}}$

$\mathbf{T(8/9) =66.67}$

Approximate

$\mathbf{T(8/9) =67}$

Hence, the soil temperature at 2:00pm is 67

Read more about composite functions at:

brainly.com/question/20379727

5 0

2 years ago

Two semicircles are attached to the sides of a rectangle as shown.

melomori [17]

Answer:

$157\ in^{2}$

Step-by-step explanation:

we know that

The area of the figure is equal to the area of rectangle plus the area of two semicircles

<u>The area of rectangle is equal to</u>

$A=14*5=70\ in^{2}$

<u>The area of the small semicircle is equal to</u>

$A=\pi r^{2} /2$

$r=5/2=2.5\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(2.5^{2})/2=9.8125 in^{2}$

<u>The area of the larger semicircle is equal to</u>

$A=\pi r^{2} /2$

$r=14/2=7\ in$ -----> radius is half the diameter

substitute

$A=(3.14)(7^{2})/2=76.93\ in^{2}$

The area of the figure is equal to

$70+9.8125+76.93=156.7425=157\ in^{2}$

8 0

3 years ago

Read 2 more answers

If f(x)= 3x squared - 2x + 1 what is f(x+2)

Crazy boy [7]

2x squared would be you final answer

4 0

3 years ago