All categories

3 years ago

Please help me with the question below and i will mark you as BRAINLIEST

Mathematics

2 answers:

d1i1m1o1n [39]3 years ago

7 0

Answer:

28.8

Step-by-step explanation:

Hope this helps!!!

zalisa [80]3 years ago

3 0

AB = 18 BC=8 CD=5

Let's call E that point at the other end of x, so x=DE.

Secant Secant theorem says

CE × CD = AC × BC

and since CE = CD + DE, AC = AB + BC

(CD + DE) CD = (AB + BC) BC

Substituting,

(5 + x)(5) = (18 + 8) (8)

5 + x = (26)(8)/5

x = -5 + (26)(8)/5 = 183/5 = 36.6

Answer: 36.6, last choice

You might be interested in

After running for 3 1/2 km, Sacha only completed half the total distance of a marathon . What is the complete distance of that m

Gre4nikov [31]

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

Answer: 7 km

Explanation:

3.5 x 2 = 7

I hope this helped!

<!> Brainliest is appreciated! <!>

- Zack Slocum

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

3 0

3 years ago

Read 2 more answers

Select the correct answer.

Rashid [163]

Answer: D. (4,3)

The x coordinates of A and B are 9 and -1 in that order. Add them up to get 9+(-1) = 9-1 = 8. Then divide by two to end up with 8/2 = 4. The midpoint has an x coordinate of 4.

Repeat for the y coordinates. Add them up: 8+(-2) = 8-2 = 6. Then divide by two: 6/2 = 3. The midpoint has an y coordinate of 3.

Those two coordinates pair up to get (x,y) = (4,3) which is the midpoint of segment AB.

7 0

3 years ago

The function h(t) = −5t2 + 15t shown in the graph models the supporting structure of a bridge:

BabaBlast [244]

Since the function is h(t) = −5t² + 15t, the domain of the function h(t) is 0 ≤ t ≤ 3

<h3>What is the domain of a function?</h3>

The domain of a function is the range of input values that make the function have real values.

Now given the function h(t) = −5t² + 15t, it will have real values when

h(t) ≥ 0

So, h(t) ≥ 0

⇒ −5t² + 15t ≥ 0

Factorizing, we have

-5t(t - 3) ≥ 0

5t(t - 3) ≤ 0

t(t - 3) ≤ 0

t ≤ 0 and t - 3 ≤ 0

t ≤ 0 and t ≤ 3

For t ≤ 0, say -1, t(t - 3) = -1(-1 - 3) = -1(-4) = 4 ≥ 0

For 0 ≤ t ≤ 3, say 2, t(t - 3) = 2(2 - 3) = 2(-1) = -2 ≤ 0

For t ≥ 3, say 4, t(t - 3) = 4(4 - 3) = 4(1) = 4 ≥ 0

Since we require t(t - 3) ≤ 0, the domain of t is thus 0 ≤ t ≤ 3

So, the domain of the function h(t) is 0 ≤ t ≤ 3

Learn more about domain of a function here:

brainly.com/question/204418

#SPJ1

8 0

2 years ago

Could someone answer this and explain an easier way to do this?

goblinko [34]

Answer:

Area of the figure = 109 cm. square

Step-by-step explanation:

The radius of the semi circle is 12 ÷ 2 = 6 cm.

Area of the semi circle = $\frac{1}{2}\pi(6^{2})=18\pi$

The radius of the quarter circle is 6 cm.

Area of the quarter circle = $\frac{1}{4}\pi(6^{2}=9\pi$

The area of the rectangle = 12 × 2 = 24

The area of the figure = 18π + 9π + 24 = 108.78 ≅ 109 cm. square

3 0

3 years ago

A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric

Murrr4er [49]

Answer:

$A = 781250\,m^{2}$ , $x = 625\,m$ , $y = 1250\,m$

Step-by-step explanation:

The perimeter covered by the electric fence in meters is:

$2\cdot x + y = 2500$

The area of the rectangle is:

$A = x\cdot y$

$A = x \cdot (2500-2\cdot x)$

Let differentiate the previous equation and equates to zero:

$2500-4\cdot x = 0$

The critical point is:

$x = 625\,m$

By the Second Derivative Text, it is proved that critical point lead to a maximum:

$\frac{d^{2}A}{dx^{2}} = -4$

The other side of the rectangle is:

$y = 1250\,m$

The largest area than can be enclosed is:

$A = (625\,m)\cdot (1250\,m)$

$A = 781250\,m^{2}$

The dimensions of the triangle are:

$x = 625\,m$

$y = 1250\,m$

3 0

4 years ago