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3 years ago

7x[(9+4)-(12-7)] please help me:(

Mathematics

2 answers:

Flauer [41]3 years ago

7 0

Answer:

7 × [ 13 - 5]

= 7 × 8

= 56

solved........

lara [203]3 years ago

3 0

Answer:

Step-by-step explanation:

7((9+4)-(12-7))

7(13-5)

7×8

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Which of the following is the measure of angle RCP?

enyata [817]

Answer:

From the information provided we have:

PD ≅ RD (= 11)

∠CPD ≅ ∠CRD (= 90°)

They both have CD as the hypotenuse.

=> ΔCPD ≅ ΔCRD

=> ∠PCD ≅ ∠RCD

Now we know that:

∠RCP = ∠PCD + ∠RCD

∠RCP = 2 · ∠RCD

∠RCP = 2 · 33° = 66°

So the answer is B

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3 years ago

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What is the lcd of 11/10 and 1/6

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2 years ago

A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a

Rashid [163]

Answer:

Length of original rectangle: 11 units.

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

$\text{Perimeter of new rectangle}=20$

Step-by-step explanation:

Let x represent width of the original rectangle.

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be $x+6$ .

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be $x-2$ .

The length of new rectangle would be $x+6-4=x+2$ .

The area of new rectangle would be $(x+2)(x-2)$ .

Now we will equate area of new rectangle with 21 and solve for x as:

$(x+2)(x-2)=21$

Applying difference of squares, we will get:

$x^2-2^2=21$

$x^2-4=21$

$x^2-4+4=21+4$

$x^2=25$

Since width cannot be negative, so we will take positive square root of both sides.

$\sqrt{x^2}=\sqrt{25}$

$x=5$

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be $x+6\Rightarrow x+5=11$ .

Therefore, the length of original rectangle is 11 units.

$\text{Area of original rectangle}=5\times 11$

$\text{Area of original rectangle}=55$

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

We know that perimeter of rectangle is two times the sum of length and width.

$\text{Perimeter of new rectangle}=2((x+2)+(x-2))$

$\text{Perimeter of new rectangle}=2((5+2)+(5-2))$

$\text{Perimeter of new rectangle}=2(7+3)$

$\text{Perimeter of new rectangle}=2(10)$

$\text{Perimeter of new rectangle}=20$

Therefore, the perimeter of the new rectangle is 20 units.

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