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3 years ago

Shkfdbhjgfdgxbvcb b c bxc

Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

6 0

Answer: 60

Step-by-step explanation: (-25/30x)-(7/30x)+(6/30x)=-26/30

-52 divided by -26/30=60

sergij07 [2.7K]3 years ago

4 0

Answer:

X = 1553/19 Hope this helped :)

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The function, y(r)=2.76r , is used to convert Russian rubles in Japanese yen, and the function, P(y)=0.16y , is used to convert

timurjin [86]

They probably want 880 pesos, since they asked you do intermediate rounding

but a more accurate answer would be
883.20 pesos

To combine the functions, you'd substitute in 2.76r for y in 0.16y. This would give you 0.16(2.76r), which simplifies to 0.4416r, but they want you round this to the nearest hundredth, so its really 0.44r. Substitute in the 2000 for r and you get 0.44 * 2000 which is 880 pesos

5 0

3 years ago

Which transformation represents a reflection over the y = x line?

dmitriy555 [2]

Answer:

see explanation

Step-by-step explanation:

Under a reflection in the line y = x

a point (x, y ) → (y, x )

4 0

3 years ago

For what values of x and y must ABCD be a parallelogram?

n200080 [17]

Step-by-step explanation:

We know that vertically opposite angles of a parallelogram are equal .

$\angle ABC =\angle ADC\ and \ \angle BAD = \angle DAC$

⇒ $3x^{\circ}=(4x-21)^{\circ} \ and \ 3y^{\circ}=(y+78)^{\circ}$

⇒ $3x^{\circ}-4x^{\circ}=-21^{\circ} \ and \ 3y^{\circ}-y^{\circ}=78^{\circ}$

⇒ $-x^{\circ}=-21^{\circ} \ and \ 2y^{\circ}=78^{\circ}$

⇒ $x^{\circ}=21^{\circ} \ and \ y^{\circ}=39^{\circ}$

7 0

3 years ago

The radius of a cylinder is 7 c m and its height is 10 c m. Find curved surface area and volume.

erma4kov [3.2K]

Answer:

CSA of the cylinder = 440 sq. cm

Volume of the cylinder = 1540 cu. cm

$\\$

Step-by-step explanation:

Given:

Radius of the cylinder = 7 cm
Height of the cylinder = 10 cm

$\\$

To Find:

Curved surface area
Volume

$\\$

Solution:

$\\$

Using formula:

$\dashrightarrow \: \: { \underline{ \boxed{ \pmb{ \sf{ \purple{CSA \: of \: cylinder = 2\pi rh}}}}}} \: \star \\ \\$

<em>Substituting the required values: </em>

$\\$

$\dashrightarrow \: \: \sf CSA {(cylinder)} = 2 \times \dfrac{22}{7} \times 7 \times 10 \\ \\ \\ \dashrightarrow \: \: \sf CSA {(cylinder)} = 2 \times 22 \times 10 \\ \\ \\ \dashrightarrow \: \: \sf CSA {(cylinder)} = 44 \times 10 \\ \\ \\ \dashrightarrow \: \: \sf CSA {(cylinder)} = 440 \: {cm}^{2} \\ \\ \\$

Now,

$\dashrightarrow \: \: { \underline{ \boxed{ \pmb{ \sf{ \purple{Volume {(cylinder)}= \pi {r}^{2} h}}}}}} \: \star \\ \\ \\$

<em>Substituting the required values, </em>

$\\$

$\dashrightarrow \: \: \sf Volume {(cylinder)}= \dfrac{22}{7} \times {(7)}^{2} \times 10 \\ \\ \\ \dashrightarrow \: \: \sf Volume {(cylinder)}= \frac{22}{7} \times 49 \times 10 \\ \\ \\ \dashrightarrow \: \: \sf Volume {(cylinder)}= 22 \times 7 \times 10 \\ \\ \\ \dashrightarrow \: \: \sf Volume {(cylinder)}= 1540 {cm}^{3} \\ \\ \\$