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3 years ago

2 pounds of peaches costed $4.20. How much will 5 pounds cost

Mathematics

2 answers:

makkiz [27]3 years ago

8 0

2 pounds is $4.20 so 1 pound is $2.10 and 5 pounds is $10.50

Vikentia [17]3 years ago

7 0

I think $8.50 is how much 5 pounds is

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Casandra is calculating how much interest she will earn in investments that are eating simple interests. She knows that simple i

Zinaida [17]

Answer:

Principal = $500

Interest = $37.50

Rate = 0.025 = 2.5%

Time = 3 years

Step-by-step explanation:

The formula for simple interest is I = PRT ÷ 100

Casandra's equation = I= 500(0.025)(3)=$37.50

Principal = $500

Interest = $37.50

Rate = 0.025 = 2.5%

Time = 3 years

3 0

3 years ago

(1+tan^2 A)cot A/cosec^2 A= tanA

Rufina [12.5K]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

1 + tan²A = sec²A , secA = $\frac{1}{cosA}$ , cosecA = $\frac{1}{sinA}$ , cotA = $\frac{cosA}{sinA}$ , tanA = $\frac{sinA}{cosA}$

Consider the left side

$\frac{(1+tan^2A)cotA}{cosec^2A}$

= $\frac{sec^2AcotA}{cosec^2A}$

= $\frac{\frac{1}{cos^2A}(\frac{cosA}{sinA}) }{cosec^2A}$

= $\frac{\frac{1}{cosAsinA} }{\frac{1}{sin^2A} }$

= $\frac{1}{cosAsinA}$ × sin²A

= $\frac{sinA}{cosA}$

= tanA = right side , thus proven

4 0

2 years ago

How do I solve radicals??

Sav [38]

Your question is too general. You might ask instead, "how do I simplify the radical sqrt(48)?"

sqrt(48) can b e simplified by rewriting 48 as the product of the perfect square 16 and the not-a-perfect-square 3:

sqrt(48) = sqrt(16)*sqrt(3) = 4*sqrt(3) (answer)

In #76, rearrange the terms so that you have 3sqrt(7x+4)=15. Then divide both sides by 3, obtaining sqrt(7x+4) =5. Squaring both sides: 7x+4=25.
Solve this linear equation for x.

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3 years ago

What is the domain of the function shown in the graph below?

Radda [10]

Answer:

can u show the graph?

Step-by-step explanation:

3 0

3 years ago

Can you help me to solve this

stealth61 [152]

Answer:

$\text{Triangle 1:}\\\\x=30\sqrt{2},\\\\\text{Triangle 2:}\\x\approx 36.18,\\?\approx12.37$

Step-by-step explanation:

*Notes (clarified by the person who asked this question):

-The triangle on the right has a right angle (angle that appears to be a right angle is a right angle)

-The bottom side of the right triangle is marked with a question mark (?)

Triangle 1 (triangle on left):

Special triangles:

In all 45-45-90 triangles, the ratio of the sides is $x:x:x\sqrt{2}$ , where $x\sqrt{2}$ is the hypotenuse of the triangle. Since one of the legs is marked as $30$ , the hypotenuse must be $\boxed{30\sqrt{2}}$

It's also possible to use a variety of trigonometry to solve this problem. Basic trig for right triangles is applicable and may be the simplest:

$\cos 45^{\circ}=\frac{30}{x},\\x=\frac{30}{\cos 45^{\circ}}=\frac{30}{\frac{\sqrt{2}}{2}}=30\cdot \frac{2}{\sqrt{2}}=\frac{60}{\sqrt{2}}=\frac{60\cdot\sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}=\frac{60\sqrt{2}}{2}=\boxed{30\sqrt{2}}\\$

Triangle 2 (triangle on right):

We can use basic trig for right triangles to set up the following equations:

$\cos 20^{\circ}=\frac{34}{x},\\x=\frac{34}{\cos 20^{\circ}}\approx \boxed{36.18}$ ,

$\tan 20^{\circ}=\frac{?}{34},\\?=34\tan 20^{\circ}\approx \boxed{12.37}$

We can verify these answers using the Pythagorean theorem. The Pythagorean theorem states that in all right triangles, the following must be true:

$a^2+b^2=c^2$ , where $c$ is the hypotenuse of the triangle and $a$ and $b$ are two legs of the triangle.

Verify $34^2+(34\tan20^{\circ})^2=\left(\frac{34}{\cos20^{\circ}}\right)^2\:\checkmark$

3 0

2 years ago