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3 years ago

Which of these quadrilateral types must have all four sides congruent?

Mathematics

2 answers:

Ronch [10]3 years ago

8 0

Answer:

Option D is correct

Step-by-step explanation:

Rectangle, square, rhombus must have all four sides congruent.

<h3>Hope it is helpful...</h3>

Paraphin [41]3 years ago

3 0

C. Square & rhombus

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Solve x-3/6 divided by 3-x/2 Please show work

Nataly_w [17]

$\cfrac{x-3}{6}: \cfrac{3-x}{2} = -\cfrac{3-x}{6}*\cfrac{2}{3-x} =- \cfrac{2}{6}=- \cfrac{1}{3}$

4 0

3 years ago

4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513

dimaraw [331]

Answer:

Part 4) $sin(\theta)=\frac{12}{13}$

Part 10) The angle of elevation is $40.36\°$

Part 11) The angle of depression is $78.61\°$

Part 12) $arcsin(0.5)=30\°$ or $arcsin(0.5)=150\°$

Part 13) $-45\°$ or $225\°$

Step-by-step explanation:

Part 4) we have that

$cos(\theta)=-\frac{5}{13}$

The angle theta lies in Quadrant II

The sine of angle theta is positive

Remember that

$sin^{2}(\theta)+ cos^{2}(\theta)=1$

substitute the given value

$sin^{2}(\theta)+(-\frac{5}{13})^{2}=1$

$sin^{2}(\theta)+(\frac{25}{169})=1$

$sin^{2}(\theta)=1-(\frac{25}{169})$

$sin^{2}(\theta)=(\frac{144}{169})$

$sin(\theta)=\frac{12}{13}$

Part 10)

Let

$\theta$ ----> angle of elevation

we know that

$tan(\theta)=\frac{85}{100}$ ----> opposite side angle theta divided by adjacent side angle theta

$\theta=arctan(\frac{85}{100})=40.36\°$

Part 11)

Let

$\theta$ ----> angle of depression

we know that

$sin(\theta)=\frac{5,389-2,405}{3,044}$ ----> opposite side angle theta divided by hypotenuse

$sin(\theta)=\frac{2,984}{3,044}$

$\theta=arcsin(\frac{2,984}{3,044})=78.61\°$

Part 12) What is the exact value of arcsin(0.5)?

Remember that

$sin(30\°)=0.5$

therefore

$arcsin(0.5)$ -----> has two solutions

$arcsin(0.5)=30\°$ ----> I Quadrant

$arcsin(0.5)=180\°-30\°=150\°$ ----> II Quadrant

Part 13) What is the exact value of $arcsin(-\frac{\sqrt{2}}{2})$

The sine is negative

The angle lies in Quadrant III or Quadrant IV

Remember that

$sin(45\°)=\frac{\sqrt{2}}{2}$

therefore

$arcsin(-\frac{\sqrt{2}}{2})$ ----> has two solutions

$arcsin(-\frac{\sqrt{2}}{2})=-45\°$ ----> IV Quadrant

$arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\°$ ----> III Quadrant

5 0

3 years ago

In the year 2003 a company made 8.8 million in profits. For each consecutive year after that their profits increased by 6%. How

yaroslaw [1]

Answer:

The company's profit in 2007 was 11.088 millions

Step-by-step explanation:

This is a compound interest problem where the initial amount is 8.8 million, the interest rate is 6% and the time period is 4 years and it gets compounded yearly. So we can use the compound interest formula, that is given by:

A = P*(1 + r/n)^(n*t)

Where A is the final amount, P is the initial amount, r is the rate, t is the total amount of time and n is the number of times it gets compounded in one year. We can now use all the values that were given to us to find out the profit of the company.

A = 8.8*(1 + (0.06))^(4) = 8.8*(1.06)^16

A = 8.8*1.26 = 11.088 millions

So the company's profit in 2007 was 11.088 millions

6 0

3 years ago

Simplify as much as possible:<br> (7nm5o2) (-3m3o5)2

Yakvenalex [24]

Answer:

63m^11 no^12

Step-by-step explanation:

Dont know if thats o or and zero but I solve it by using o..if it was a zero then comment that so I can redo it anyways heres the explanation

(7nm^5 o^2) × (-3m^3 o^5)^2

7m^5 no^2) ×(-3m^3 o^5) ^2

(7m^5 no^2) × (3m^3 o^5)^2

7m^5 no^2 × (3m^3 o^5) ^2

7m^5 no^2 × 9m^6 o^10

63m^11 no^12

3 0

3 years ago

What’s the answer to this problem that I can not figure out

Maslowich

Answer:

-64/3 HOPE THIS HELPS if you need more help let me know

4 0

3 years ago