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2 years ago

Hi everyone same as a while ago please help me with this :))) (with solution plzz) Thank you!

Mathematics

2 answers:

Irina18 [472]2 years ago

7 0

Answer:

Hello there

Your answer is

6. 8.2g7. 3800 ml8. 560 L9. 6300 ml10. 30.48cm11. 21120 ft12. 487.48cm

Step-by-step explanation:

820 ÷ 100 3.8 × 100056000 ÷ 10063 × 100 12 × 2.544 × 528016 × 30.48

hope it’s helps you Buddy

~Zayn

solniwko [45]2 years ago

6 0

Answer:

6. 8.2g

7. 3800 ml

8. 560 L

9. 6300 ml

10. 30.48cm

11. 21120 ft

12. 487.48cm

Step-by-step explanation:

820 ÷ 100

3.8 × 1000

56000 ÷ 100

63 × 100

12 × 2.54

4 × 5280

16 × 30.48

Please mark me as brainliest

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Please answer this correctly

suter [353]

Answer:

1 11/24 tablespoons

Step-by-step explanation:

$1+1+1\frac{1}{3} +1\frac{1}{3} +1\frac{2}{3} +1\frac{2}{3} +1\frac{2}{3} +2=\\\\2+2\frac{2}{3} +3\frac{6}{3} +2=\\\\9\frac{8}{3} =\\\\11\frac{2}{3}$

There were 8 mugs in total so:

$11\frac{2}{3}$ ÷ 8=

$11\frac{2}{3} *\frac{1}{8} =\\\\\frac{35}{3} *\frac{1}{8} =\\\\\frac{35}{24} =\\\\1\frac{11}{24}$

5 0

3 years ago

Paul Havlik promised his grandson Jamie that he would give him $7,100 7 years from today for graduating from high school. Assume

Dominik [7]

Answer:

$\large \boxed{\$4100.07}$

Step-by-step explanation:

The formula for the future value (FV) of an investment earning compound interest is

$FV = PV \left (1 + \frac{r}{n} \right )^{nt}$

where

PV = the present value (PV) of the money invested

r = the annual interest rate expressed as a decimal fraction

t = the time in years

n = the number of compounding periods per year

Data:

FV = $7100

r = 8 % = 0.08

t = 7 yr

n = 2

Calculation:

$\begin{array}{rcl}\\7100& =& PV \left (1 + \dfrac{0.08}{2} \right )^{2 \times 7}\\\\& =& PV (1 + 0.04)^{14}\\\\& =&PV (1.04)^{14}\\& =& PV(1.731676)\\PV& =& \dfrac{7100}{1.731676}\\\\& =& \mathbf{4100.07}\\\end{array}\\\text{The present value of the money is $\large \boxed{\mathbf{\$4100.07}}$}$

4 0

2 years ago

Which criterion can be used to prove abc and adc are congruent? triangles, abc and acd, are plotted on a grid that share a commo

RUDIKE [14]

The correct answer is option D which is ASA ( Angle - Side - Angle).

<h3>What is congruency?</h3>

If two triangles are congruent, this means they are the same shape and the same size. By definition, all corresponding angles of two congruent triangles are congruent. Additionally, all corresponding sides of two congruent triangles are congruent.

In triangles ABC and ADC,

Angles BCA and DAC are congruent (from the diagram);

Angles BAC and DCA are congruent (from the diagram);

AC is congruent AC (by reflexive property).

So, the triangles ABC and CDA are congruent with the ASA postulate.

ASA (Angle-Side-Angle) postulate states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.

Therefore the correct answer is option D which is ASA ( Angle - Side - Angle).

To know more about congruency follow

brainly.com/question/2938476

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4 0

1 year ago

Round 427,266 to the nearest ten thousand

melamori03 [73]

Answer:

430,000

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

The Cartesian coordinates of a point are given. (a) (−3, 3) (i) Find polar coordinates (r, θ) of the point, where r > 0 and 0

irina [24]

Answer:

a) (-3, 3)

(i) Polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ)

= (3√2, 0.75π)

(ii) Polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ)

= (-3√2, 1.75π)

b) (4, 4√3)

(i) Polar coordinates (r, θ) of the point, where r > 0 and 0 ≤ θ < 2π. (r, θ)

= (8, 0.13π)

(ii) Polar coordinates (r, θ) of the point, where r < 0 and 0 ≤ θ < 2π. (r, θ)

= (-8, 1.13π)

Step-by-step explanation:

We know that polar coordinates are related to (x, y) coordinates through

x = r cos θ

y = r sin θ

And r = √[x² + y²]

a) For (-3, 3)

(i) x = -3, y = 3

r = √[x² + y²] = √[(-3)² + (3)²] = √18 = ±3√2

If r > 0, r = 3√2

x = r cos θ

-3 = 3√2 cos θ

cos θ = -3 ÷ 3√2 = -(1/√2)

y = r sin θ

3 = 3√2 sin θ

sin θ = 3 ÷ 3√2 = (1/√2)

Tan θ = (sin θ/cos θ) = -1

θ = 0.75π or 1.75π

Note that although, θ = 0.75π and 1.75π satisfy the tan θ equation, only the 0.75π satisfies the sin θ and cos θ equations.

So, (-3, 3) = (3√2, 0.75π)

(ii) When r < 0, r = -3√2

x = r cos θ

-3 = -3√2 cos θ

cos θ = -3 ÷ -3√2 = (1/√2)

y = r sin θ

3 = -3√2 sin θ

sin θ = 3 ÷ -3√2 = -(1/√2)

Tan θ = (sin θ/cos θ) = -1

θ = 0.75π or 1.75π

Note that although, θ = 0.75π and 1.75π satisfy the tan θ equation, only the 1.75π satisfies the sin θ and cos θ equations.

So, (-3, 3) = (-3√2, 1.75π)

b) For (4, 4√3)

(i) x = 4, y = 4√3

r = √[x² + y²] = √[(4)² + (4√3)²] = √64 = ±8

If r > 0, r = 8

x = r cos θ

4 = 8 cos θ

cos θ = 4 ÷ 8 = 0.50

y = r sin θ

4√3 = 8 sin θ

sin θ = 4√3 ÷ 8 = (√3)/2

Tan θ = (sin θ/cos θ) = (√3)/4

θ = 0.13π or 1.13π

Note that although, θ = 0.13π and 1.13π satisfy the tan θ equation, only the 0.13π satisfies the sin θ and cos θ equations.

So, (4, 4√3) = (8, 0.13π)

(ii) When r < 0, r = -8

x = r cos θ

4 = -8 cos θ

cos θ = 4 ÷ -8 = -0.50

y = r sin θ

4√3 = -8 sin θ

sin θ = 4√3 ÷ -8 = -(√3)/2

Tan θ = (sin θ/cos θ) = (√3)/4

θ = 0.13π or 1.13π

Note that although, θ = 0.13π and 1.13π satisfy the tan θ equation, only the 1.13π satisfies the sin θ and cos θ equations.

So, (4, 4√3) = (-8, 1.13π)

Hope this Helps!!!

8 0

3 years ago