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Vadim26 [7]
3 years ago
10

A flagpole has a height of 2222 yards. It will be supported by three​ cables, each of which is attached to the flagpole at a poi

nt 66 yards below the top of the pole and attached to the ground at a point that is 1212 yards from the base of the pole. Find the total number of yards of cable that will be required.
Mathematics
1 answer:
mojhsa [17]3 years ago
7 0
Nvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
You might be interested in
Look down below for question
dusya [7]
So x=x
and y=pounds
when x=0, y=64
when x=1, y=63.1
to find slope, (y1-y2)/(x1-x2)=(64-63.1)/(0-1)=0.9/-1=-0.9=slope
y=-0.9x+b
subsitute
64=-0.9(0)+b
b=64
y=-0.9+64
so int 2001, x=11
y=-0.9(11)+64
y=-9.9+64
y=54.1
5 0
3 years ago
Cynthia is in the process of constructing two tangent line segments from a point outside circle O to circle O as shown. Which li
Allisa [31]

Answer:

(B) Segments MA and MB

Step-by-step explanation:

The tangent to the circle at a point is perpendicular to the radius of the circle drawn to the point of tangency.

Tangent at a point is unique.

Since there can be no two tangents at a point on circle, the options (b) and (c) are ruled out.

Now, if OA is perpendicular to MA, MA is the tangent else if OA is perpendicular to PA, PA is the tangent. Same is the case with point B.

Tangents from the same external point has same length.

MA = MB since they are the radii of the same circle with center M.

Hence, MA and MB meet all the requirements of the tangents.


8 0
4 years ago
Read 2 more answers
Can anyone do 4,5 and 6 for me plz
cricket20 [7]

For  question 4, sinA =\frac{a}{c}, cosA= \frac{b}{c}, tan B = \frac{b}{a}, sin J = \frac{j}{l}, cosK = \frac{j}{l}, tanK = \frac{k}{j}.

Question 5. Option a and question 6. Option j

Step-by-step explanation:

Step 1:

The three basic formula needed to solve these questions are:

sin\theta = \frac{oppositeside}{hypotenuse} , cos\theta = \frac{adjacentside}{hypotenuse}, tan\theta= \frac{opposite side}{adjacent side}.

Step 2:

Using the above formula, we solve the following values

sinA = \frac{oppositeside}{hypotenuse}  =\frac{a}{c}.

cosA = \frac{adjacentside}{hypotenuse} = \frac{b}{c}.

tanB= \frac{opposite side}{adjacent side}= \frac{b}{a}.

sinJ = \frac{oppositeside}{hypotenuse} = \frac{j}{l}.

cosK = \frac{adjacentside}{hypotenuse}= \frac{j}{l}.

tanK= \frac{opposite side}{adjacent side}= \frac{k}{j}.

Step 3:

For question 5, The triangle's angle = 23°, opposite side = BC inches and hypotenuse = 4 inches.

sin\theta= \frac{opposite side}{hypotenuse}. sin 23^{\circ}= \frac{BC}{4}, sin23^{\circ} = 0.3907,BC = (0.3907)(4) = 1.5628.

SO BC is 1.5628 inches, rounding this off to the nearest tenth, we get BC = 1.6 inches which is option a.

Step 4:

For question 6, The triangle's angle = 50°, opposite side = QR m the adjacent side = 8.1 m.

tan\theta= \frac{opposite side}{adjacentside}. tan 50^{\circ}=\frac{QR}{8.1}, tan50^{\circ} = 1.1917,QR = (1.1917)(8.1) = 9.65277

SO QR is 9.65277 meters, rounding this off to the nearest tenth, we get QR = 9.7 inches which is option j.

7 0
3 years ago
Laqueta has three times as many quarters as Pablo if they each spend $.50 Laqueta will have five times as many quarters as Pablo
masya89 [10]

Answer:

Number of quarters with Pablo = 4

Number of quarters with Laqueta = 12

Step-by-step explanation:

Given that:

Laqueta has three times as many quarters as Pablo.

They both spend $0.50.

After that Laqueta will have 5 times as many quarters as Pablo.

To find:

Number of quarters with Laqueta and Pablo now?

Solution:

Let number of quarters with Laqueta = x

Let number of quarters with Pablo = y

As per given statement:

x=3y ..... (1)

In $0.50, there are 2 quarters:

After spending $0.50, number of quarters left with both of them:

Number of quarters left with Laqueta = x - 2 =  3y-2

Money left with Pablo = y - 2

As per given statement:

3y-2 = 5(y-2)\\\Rightarrow 3y-2 = 5y-10\\\Rightarrow 2y = 8\\\Rightarrow y = 4

Therefore, the answer is:

Number of quarters with Pablo = 4

Number of quarters with Laqueta = 3 \times 4 = 12

8 0
3 years ago
What is another way to write the expression ​ t⋅(14−5) ​?
grigory [225]

t⋅(14−5)

distribute

14t -5t

Choice A

8 0
3 years ago
Read 2 more answers
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