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3 years ago

dan eat 3 boxes of jellybeans. if the re are 12 jellybeans in wach box , how mean jellybeans did dan eat in all

Mathematics

2 answers:

rjkz [21]3 years ago

8 0

Multiply twelve by three and the ans is 36

VikaD [51]3 years ago

7 0

Dan eat all jelly beans he am many hungry?

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A man places a mirror on the ground and sees the reflection of the top of a flagpole as shown in the figure below. The two trian

tigry1 [53]

Answer:

Step-by-step explanation:

5 0

3 years ago

The value of <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20-%2025%7D%20" id="TexFormula1" title=" \sqrt{ - 25} " alt=" \

Andreas93 [3]

2.23606797749979
This is the answer

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2 years ago

S 8,000 a good estimate for the sum of 3,816 and 467?

Damm [24]

Answer:

it is not a good estimate.

Step-by-step explanation:3816 can be rounded to 4000 and 467 can be rounded to 500. Adding 4000+500 gives 4500, which isn't near 8000.

i hoped this helped :)

7 0

3 years ago

The volume V of a solid right circular cylinder is given by V = πr2h where r is the radius of the cylinder and h is its height.

emmainna [20.7K]

Answer:

8.20in³

Step-by-step explanation:

Given V = πr²h

r is the radius = 1.5in

h is the height = 6in

thickness of wall of the cylinder dr = 0.04in

top and bottom thickness dh 0.07in+0.07in = 0.14in

To compute the volume, we will find the value of dV

dV = dV/dr • dr + dV/dh • dh

dV/dr = 2πrh

dV/dh = πr²

dV = 2πrh dr + πr² dh

Substituting the values into the formula

dV = 2π(1.5)(6)•(0.04) + π(1.5)²(6) • 0.14

dV = 2π (0.36)+π(1.89)

dV = 0.72π+1.89π

dV = 2.61π

dV = 2.61(3.14)

dV = 8.1954in³

Hence volume, in cubic inches, of metal in the walls and top and bottom of the can is 8.20in³ (to two dp)

7 0

3 years ago

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago