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3 years ago

Which equation can be used to find the measure of angle BAC?

Mathematics

2 answers:

asambeis [7]3 years ago

8 0

Answer:

Step-by-step explanation:

let angle $B$ . be $y$ then,

using angle sum property,

x+y+90=180 (angle c is 90 because its given)

hence you have the equation

Monica [59]3 years ago

6 0

Answer:

Cos() = AC / AB

So in this case: Cos() = 5 / 13

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12345 [234]

5 h
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10 28

You would cross multiply and get 140=10h and you divide both sides by 10 and get 14=h. So the flagpole is 14 feet tall.

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3 years ago

The crayola crayon company can make 2400 crayons in 4 minutes.How many crayons can they make in 15 minuttes

scZoUnD [109]

4 x n = 2400 x 15Find the cross products
4n = 36000Multiply
n = 9000Divide both sides by 4

The Crayola crayon company can make 9000 crayons in 15 minutes.

hope this helps :)

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3 years ago

PlEASE HELP ME PEOPLE HUMPTH OK ILL SAY IT PEOPLE WANT MONEY TO ANSWER PROBLEMS HERE If the mean of six nu m b e rs is 48, does

SpyIntel [72]

Answer:

No, none of the number need to be 48 for the mean to be 48. To get a mean, you add up all the number and divide it by the amount of numbers.

Example:

the mean of 10, 79, 42, 88, 19, and 50 is 48, but the actual number 48 was not part of the set.

10 + 79 + 42 + 88 + 19 + 50 = 288

288 ÷ 6 = 48

6 0

3 years ago

A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six

kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

$V = Length \times Width \times Height \\\\$

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

$Width = 3 - 2x \\\\$

The length of the shaded region is given by

$Length = \frac{1}{2} (5 - 3x) \\\\$

So, the volume of the box becomes,

$V = \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V = \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V = \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V = \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\$

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

$\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\$

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

$a = 18 \\\\b = -38 \\\\c = 15 \\\\$

$x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 + 19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\$