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2 years ago

12

the triangular opening at the front of the tent is 8 feet at the bottom. The sides are 10 feet long each. How tall is the tent a

t the top peak of the opening?

1 answer:

kkurt [141]2 years ago

5 0

I believe that it is square root 84 feet

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A tangent and a secant x = ?

guajiro [1.7K]

x = 45°

Solution:

Given data:

Measure of larger arc = 152°

Measure of smaller arc = 62°

<em>If a tangent and a secant intersect at the exterior of a circle then the measure of angle formed is one-half the positive difference of the measures of the intercepted arcs.</em>

$$\Rightarrow x = \frac{1}{2}(\text{ larger arc }- \text{smaller arc} )$

$$\Rightarrow x = \frac{1}{2}(152^\circ-62^\circ)$

$$\Rightarrow x = \frac{1}{2}(90^\circ)$

⇒ x = 45°

The value of x is 45°.

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2 years ago

In mrs.smiths class every time you turn in 10 assignments you earn 1 homework pass.Write 3 equivalent ratios if turned in assign

frutty [35]

Answer:

Step-by-step explanation:

For every 10 assignments you get 1 homework pass. So you would multiply them by two and up more.

So here's some of your answers:

10:1

20:2

30:3

40:4

50:5

60:6

70:7

80:8

90:9

100:10

You would just continue this pattern.

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2 years ago

Sharina simplified the expression 3(2x – 6 – x + 1)2 – 2 + 4x. In Step 1 she simplified within the parentheses. In Step 2 she ex

stiks02 [169]

3x^2 - 42x + 147 - 2 + 4x

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2 years ago

Marcus states that the polynomial expression 3x3 – 4x2y + y2 + 2 is in standard form. Ariel states that it should be y2 – 4x2y +

lubasha [3.4K]

The degree of a polynomial is the highest power of its terms.
The power of a term is the sum of the powers of all the variables in a term.
A polynomial is written starting with the greatest power in standard form.
In the first case, the power of the first term is 3, the power of the second is 3 (2 from x + 1 from y) but the power of x has decreased so it is the second term, and then so on.
In the second case, the power is starting form 2 and then increasing to 3. This is incorrect.
Therefore, Marcus' suggestion is correct.

4 0

2 years ago

Read 2 more answers

An Archer shoots an arrow horizontally at 250 feet per second. The bullseye on the target and the arrow are initially at the sam

Genrish500 [490]

Answer:

<h2>1.84feet</h2>

Step-by-step explanation:

Using the formula for finding range in projectile, Since range is the distance covered in the horizontal direction;

Range $R = U\sqrt{\frac{H}{g} }$

U is the velocity of the arrow

H is the maximum height reached = distance below the bullseye reached by the arrow.

R is the horizontal distance covered i.e the distance of the target from the archer.

g is the acceleration due to gravity.

Given R = 60ft, U = 250ft/s, g = 32ft/s H = ?

On substitution,

$60 = 250\sqrt{\frac{H}{32}} \\\frac{60}{250} = \sqrt{\frac{H}{32}}\\\frac{6}{25} = \sqrt{\frac{H}{32}$

Squaring both sides we have;

$(\frac{6}{25} )^{2} = (\sqrt{\frac{H}{32} } )^{2} \\\frac{36}{625} = \frac{H}{32} \\625H = 36*32\\H = \frac{36*32}{625} \\H = 1.84feet$

The arrow will hit the target 1.84feet below the bullseye.

5 0

2 years ago