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2 years ago

Which two of the objects shown below could we slice diagonal to their bases/faces to create ellipse cross-

Mathematics

1 answer:

nalin [4]2 years ago

5 0

Answer: cone and cylinder

Step-by-step explanation: idk but it’s the right answers

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Find the appropriate perimeter of the figure below

yanalaym [24]

Answer:

55.7 in.

Step-by-step explanation:

I assume the figure is a rectangle and a half circle.

The three sides of the rectangle that need to be added toward the perimeter measure 15 in., 10 in., and 15 in.

Then you need the circumference of the half circle which has a 10-inch diameter.

circumference = (pi)d

circumference = 3.14 * 10 in. = 31.4 in.

We only need half the circumference, so we have 31.4 in./2 = 15.7 in.

Now we add the three sides of the rectangle listed above and the circumference of the half circle.

perimeter = 15 in. + 10 in. + 15 in. + 15.7 in.

perimeter = 55.7 in.

8 0

2 years ago

Read 2 more answers

PLEASE HELP!

IrinaVladis [17]

Answer:

Step-by-step explanation:

<h3>Sphere:</h3>

1) a) r = 7 cm

$\sf \boxed{\text{Volume of sphere=$\dfrac{4}{3}\pi r^3$}}$

$\sf =\dfrac{4}{3}*3.14*7*7*7\\\\= 1436 \ cm^3$

$\sf \boxed{\text{\bf Surface area of sphere = $4\pi r^2$}}$

= 4*3.14 * 7 * 7

= 615.44 cm²

b) r= 8.4 cm

$Volume = \dfrac{4}{3}\pi r^3$

$\sf =\dfrac{4}{3}*\dfrac{22}{7}*8.4*8.4*8.4\\\\= 2483.7 \ cm^3$

Surface area = 4*3.14*8.4*8.4

= 887.04 cm²

<h3>Hemisphere:</h3>

a) r = 6 cm

$\sf \boxed{\text{\bf Volume of hemisphere =$\dfrac{2}{3} \pi r^3$}}$

$\sf =\dfrac{2}{3}*3.14*6*6*6\\\\= 452.16 \ cm^3$

$\sf \boxed{\text{\bf Surface area of hemisphere=3 \pi r^2}}$ $\sf \boxed{\text{\bf Surface area of hemisphere = $3 \pi r^2$}}$

= 3 * 3.14 * 6 * 6

= 339.12 cm²

3 0

1 year ago

Use the perfect square method <br> (3x + 3)^2 = 36

kkurt [141]

Factor out the common term; 3

(3(x + 1))^2 = 36

Use the Multiplication Distributive Property; (xy)^a = x^ay^a

3^2(x + 1)^2 = 36

Simplify 3^2 to 9

9(x + 1)^2 = 36

Divide both sides by 9

(x + 1)^2 = 36/9

Simplify 36/9 to 4

(x + 1)^2 = 4

Take the square root of both sides

x + 1 = √4

Since 2 * 2 = 4, the square root of 2 is 2

x + 1 = 2

Break down the problem into these 2 equations

x + 1 = 2

x + 1 = -2

Solve the first equation; x + 1 = 2

x = 1

Solve the second equation; x + 1 = -2

x = -3

Collect all solutions;

8 0

2 years ago

What are the potential solutions of log4x+log4(x+6)=2?

lions [1.4K]

The potential solutions of $log_4x+log_4(x+6)=2$ are 2 and -8.

<h3>Properties of Logarithms</h3>

From the properties of logarithms, you can rewrite logarithmic expressions.

The main properties are:

Product Rule for Logarithms - $log_{b}(a*c)=log_{b}a+log_{b}c$
Quotient Rule for Logarithms - $log_{b}(\frac{a}{c} )=log_{b}a-log_{b}c$
Power Rule for Logarithms - $log_{b}(a^c)=c*log_{b}a$

The exercise asks the potential solutions for $log_4x+log_4(x+6)=2$ . In this expression you can apply the Product Rule for Logarithms.

$log_4x+log_4(x+6)=2\\ \\ x*(x+6)=4^2\\ \\ x^2+6x=16\\ \\ x^2+6x-16=0$

Now you should solve the quadratic equation.

Δ= $b^2-4ac=36-4*1*(-16)=36+64=100$ . Thus, x will be $x_{1,\:2}=\frac{-6\pm \:\sqrt{100} }{2\cdot \:1}=\frac{-6\pm \:10}{2}$ . Then:

$x_1=\frac{-6+10}{2}=\frac{4}{2} =2\\ \\ \:x_2=\frac{-6-10}{2}=\frac{-16}{2} =-8$

The potential solutions are 2 and -8.

Read more about the properties of logarithms here:

brainly.com/question/14868849

4 0

1 year ago

PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS

Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

$a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\$

With respect to reference angle A, we have:

opposite side = a = 4
adjacent side = b = $\sqrt{33}$
hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\$

Then cosine which is adjacent over hypotenuse

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\$

Tangent is the ratio of opposite over adjacent

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\$

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

cosecant, abbreviated as csc, is the reciprocal of sine
secant, abbreviated as sec, is the reciprocal of cosine
cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

$\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{ ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

------------------------------------------------------

Summary:

The missing side is $b = \sqrt{33}$

The 6 trig functions have these results

$\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0

10 months ago