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2 years ago

Quiz

Mathematics

1 answer:

yan [13]2 years ago

7 0

Answer:

leaving each unit price as an unrounded decimal

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Referring to the Fig. in Question #40, find the unknown length Y.

icang [17]

Answer:

tan30°=7/x

$\frac{1}{ \sqrt{3} } = \frac{7}{x } \\ x = 7 \sqrt{3}$

cos30°=x/y

$\frac{ \sqrt{3} }{2} = \frac{7 \sqrt{3} }{y} \\ \\ 7 \sqrt{3} \times 2 = \sqrt{3} \times y \\ \frac{7 \sqrt{3 } \times 2}{ \sqrt{3} } = y \\7 \times 2 = y \\ 14 = y$

tell me if it's wrong.

7 0

2 years ago

Read 2 more answers

What is the value of z?

babymother [125]

Answer:

Step-by-step explanation:

Item 1:

Discount = Original price - discount price

= 22.50 - 13.50

= $ 9

Discount percentage = $\frac{9}{22.50}*100\\\\$

= 40%

Discount percentage = 40%

Tax = 14.31 - 13.50 = $0.81

Tax% = $\frac{0.81}{13.50}*100\\\\$ = 6%

Tax percentage = 6%

Item 2:

Tax = 6% of 16.20

= 0.06 * 16.20

= 0.972

Tax = $ 0.97

Total price (including tax )= 16.20 + 0.97 = $ 17.17

x = $ 17.17

Item 3:

Discount = 40% of 32.50

= 0.4 * 32.50

= $ 13

y = 32.50 -13

y = $ 19. 50

Tax = 6% of 19.50

= 0.06 * 19.50

= 1.17

z = 1.17 + 19.50

Z =$ 20.67

5 0

2 years ago

Eight more than the quotient of a number and 5 is 3

spin [16.1K]

The answer is -25
3-8=-5
-5*5=-25

7 0

2 years ago

Find the slope of a line perpendicular to 2x-y=16

morpeh [17]

The slope would be -1/2

5 0

3 years ago

Read 2 more answers

Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:

Papessa [141]

Answer: it is verified that:

* y1 and y2 are solutions to the differential equation,

* c1 + c2t^(1/2) is not a solution.

Step-by-step explanation:

Given the differential equation

yy'' + (y')² = 0

To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.

Now,

y1 = 1

y1' = 0

y'' = 0

So,

y1y1'' + (y1')² = (1)(0) + (0)² = 0

Hence, y1 is a solution.

y2 = t^(1/2)

y2' = (1/2)t^(-1/2)

y2'' = (-1/4)t^(-3/2)

So,

y2y2'' + (y2')² = t^(1/2)×(-1/4)t^(-3/2) + [(1/2)t^(-1/2)]² = (-1/4)t^(-1) + (1/4)t^(-1) = 0

Hence, y2 is a solution.

Now, for some nonzero constants, c1 and c2, suppose c1 + c2t^(1/2) is a solution, then y = c1 + c2t^(1/2) satisfies the differential equation.

Let us differentiate this twice, and verify if it satisfies the differential equation.

y = c1 + c2t^(1/2)

y' = (1/2)c2t^(-1/2)

y'' = (-1/4)c2t(-3/2)

yy'' + (y')² = [c1 + c2t^(1/2)][(-1/4)c2t(-3/2)] + [(1/2)c2t^(-1/2)]²

= (-1/4)c1c2t(-3/2) + (-1/4)(c2)²t(-3/2) + (1/4)(c2)²t^(-1)

= (-1/4)c1c2t(-3/2)

≠ 0

This clearly doesn't satisfy the differential equation, hence, it is not a solution.

6 0

2 years ago