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2 years ago

7

Morgan determined he needed 45 1/2 yards of wire fencing for his landscaping project. He already had 17 3/4 yards. How much more

does he need to purchase?

1 answer:

gregori [183]2 years ago

7 0

The answer is 27 3/4

You might be interested in

A rectangular patio has a length of 12 1\2 feet and an area of 103 1\8 square feet. What is the width of the patio?

Whitepunk [10]

Answer:

8.25 feet

Step-by-step explanation:

to find the width, you divide the area by the length

103 1/8 / 12 1/2 = 8.25

Give brainliest, please!
Hope this helps :)

8 0

2 years ago

A large square consists of four identical rectangles and a small square. The area of the larg square is 49 cm2 and the length of

trapecia [35]

Answer:

The area of the small square is 1 cm^2

Step-by-step explanation:

The large square consist in four identical rectangles and one small square.

Then the area of the small square will be equal to the difference between the area of the large square and the areas of the rectangles.

Because we have 4 equal rectangles, if R is the area of one rectangle, and S is the area of the large square, the area of the small square will be:

area = S - 4*R

We know that the area of the large square is 49 cm^2

Then:

S = 49cm^2

Remember that the area of a square of side length K is:

A = K^2

Then the side length of the large square is:

K^2 = 49 cm^2

K = √(49 cm^2) = 7cm

And we know that the diagonal of one rectangle is 5cm.

Remember that for a rectangle of length L and width W, the diagonal is:

D = √(L^2 + W^2)

Then:

D = √(L^2 + W^2) = 5cm

And for how we construct this figure, we must have that the length of the rectangle plus the width of the rectangle is equal to the side length of the large square, then:

L + W = 7cm

L = (7cm - W)

Replacing this in the diagonal equation, we get:

√((7cm - W)^2 + W^2) = 5cm

(7cm - W)^2 + W^2 = (5cm)^2 = 25cm^2

49cm^2 - 14cm*W + W^2 + W^2 = 25cm^2

2*W^2 - 14cm*W + 49cm^2 = 25cm^2

2*W^2 - 14cm*W + 49cm^2 - 25cm^2 = 0

2*W^2 - 14cm*W + 24cm^2 = 0

We can solve this for W using the Bhaskara's formula, the solutions are:

$W = \frac{-(-14cm) \pm \sqrt{(-14cm)^2 - 4*2*(24cm^2)} }{2*2} = \frac{14cm \pm 2cm}{4}$

Then we have two solutions, and we only need one (because the length will have the other value)

We can take:

W = (14 cm + 2cm)/4 = 4cm

Then using the equation:

L + W = 7cm

L + 4cm = 7cm

L = 7cm - 4cm = 3cm

L = 3cm

Now remember that the area of one rectangle of length L and width W is:

R = L*W

Then the area of one of these rectangles is:

R = 4cm*3cm = 12cm^2

Now we can compute the area of the small square:

area = S - 4*R = 49cm^2 - 4*12cm^2 = 1cm^2

The area of the small square is 1 cm^2

3 0

3 years ago

M=2 (-1, -4) slope intersect form

aalyn [17]

$y=mx+b\\\\
-4=2\cdot(-1)+b\\
-4=-2+b\\
b=-2\\\\
\boxed{y=2x-2}$

8 0

3 years ago

A balance scale was in perfect balance when Jane placed a box of candy on one pane of the balance and 3/4 of the same sized cand

dangina [55]

Answer:

Step-by-step explanation:

i'm pretty sure that you will have to add it.

so, when we add two fractions such as 3/4 + 3/4, we make sure that the denominators (the bottom numbers) are the same and then we simply add the numerators (the top numbers).

in this problem, the denominators are the same so we will simply add 3+3 which equals to 6/4. the denominator will remain the same.

<em>answer:</em>

3/4 + 3/4 = 6/4

<u>i hope this is helpful. </u>

5 0

3 years ago

Find the domain and range of f(x). Please explain your answer.<br> f(x) = √(x^2+5x+6)

timama [110]

Answer:

Domain:— [ x ≥ -2, x ≤ -3 ]

Range:— [ y ≥ 0 ]

Step-by-step explanation:

You may use graphing calculator to draw a graph and examine the graph’s domain and range. However, I’ll explain further about the graph of quadratic in a surd.

First, factor the quadratic expression in the surd:—

$\displaystyle \large{f(x)=\sqrt{(x+3)(x+2)}}$

We can find the x-intercepts by letting f(x) = 0.

I’ll be separating in two parts — one for finding x-intercept and one for finding y-intercept.

__________________________________________________________

Finding x-intercepts

Let f(x) = 0.

$\displaystyle \large{0=\sqrt{(x+3)(x+2)}}$

Solve for x, square both sides:—

$\displaystyle \large{0^2 = (\sqrt{(x+3)(x+2)})^2}\\\displaystyle \large{0=(x+3)(x+2)}$

Simply solve a quadratic equation:—

$\displaystyle \large{x=-3,-2}$

Therefore, x-intercepts are:—

$\displaystyle \large{\boxed{x=-3,-2}}$

__________________________________________________________

Finding y-intercept

Let x = 0.

$\displaystyle \large{f(x)=\sqrt{(0+3)(0+2)}}\\\displaystyle \large{f(x)=\sqrt{3 \cdot 2}}\\\displaystyle \large{f(x)=\sqrt{6}}$

Therefore, y-intercept is:—

$\displaystyle \large{\boxed{\sqrt{6}}}$

__________________________________________________________

However, I want you to focus on x-intercepts instead. We know that the square root only gives you a positive value. That means the range of function can only be y ≥ 0.

For domain, first, we have to know how or what the graph looks like. You can input the function in a graphing calculator as you’ll see that when x ≥ -2, the graph heads to the right while/when x ≤ -3, the graph heads to the left. This means that the lesser value of x-intercept gets left and more value get right.

See, between -3 < x < -2, there is no curve, point or anything between the interval. Therefore, -3 < x < -2 does not exist in function.

Hence, the domain is:—

x ≥ -2, x ≤ -3

4 0

2 years ago