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3 years ago

6

: 2x(5x + 3) + 7(5x + 3)

1 answer:

Setler [38]3 years ago

7 0

10x2 + 6x + 35x + 21

10x2 + 41x + 21

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50 points! Please help mhanifa I need this done fast! If anyone else answers with the wrong answer you will be reported

Black_prince [1.1K]

Answer:

Yes it can because an isoseles triangle is just a triangle with two equale sides

Step-by-step explanation:

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3 years ago

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Find the new amount given the original amount and the percent of change. $5; 40% increase The new amount is $

bekas [8.4K]

Answer:

3

Step-by-step explanation:

5$ -40%=3

7 0

3 years ago

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Determine what shape is formed for the given coordinates for ABCD, and then find the perimeter and area as an exact value and ro

Helga [31]

Answer:

Part 1) The shape is a trapezoid

Part 2) The perimeter is $25(4+\sqrt{2})\ units$ or approximately $135.4\ units$

Part 3) The area is $937.5\ units^2$

Step-by-step explanation:

step 1

Plot the figure to better understand the problem

we have

A(-28,2),B(-21,-22),C(27,-8),D(-4,9)

using a graphing tool

The shape is a trapezoid

see the attached figure

step 2

Find the perimeter

we know that

The perimeter of the trapezoid is equal to

$P=AB+BC+CD+AD$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance AB

we have

A(-28,2),B(-21,-22)

substitute in the formula

$d=\sqrt{(-22-2)^{2}+(-21+28)^{2}}$

$d=\sqrt{(-24)^{2}+(7)^{2}}$

$d=\sqrt{625}$

$d_A_B=25\ units$

Find the distance BC

we have

B(-21,-22),C(27,-8)

substitute in the formula

$d=\sqrt{(-8+22)^{2}+(27+21)^{2}}$

$d=\sqrt{(14)^{2}+(48)^{2}}$

$d=\sqrt{2,500}$

$d_B_C=50\ units$

Find the distance CD

we have

C(27,-8),D(-4,9)

substitute in the formula

$d=\sqrt{(9+8)^{2}+(-4-27)^{2}}$

$d=\sqrt{(17)^{2}+(-31)^{2}}$

$d=\sqrt{1,250}$

$d_C_D=25\sqrt{2}\ units$

Find the distance AD

we have

A(-28,2),D(-4,9)

substitute in the formula

$d=\sqrt{(9-2)^{2}+(-4+28)^{2}}$

$d=\sqrt{(7)^{2}+(24)^{2}}$

$d=\sqrt{625}$

$d_A_D=25\ units$

Find the perimeter

$P=25+50+25\sqrt{2}+25$

$P=(100+25\sqrt{2})\ units$

simplify

$P=25(4+\sqrt{2})\ units$ ----> exact value

$P=135.4\ units$

therefore

The perimeter is $25(4+\sqrt{2})\ units$ or approximately $135.4\ units$

step 3

Find the area

The area of trapezoid is equal to

$A=\frac{1}{2}[BC+AD]AB$

substitute the given values

$A=\frac{1}{2}[50+25]25=937.5\ units^2$

4 0

3 years ago

If sin ⁡x=725, and 0 ∠ x ∠ pi/2, what is the tan (x - pi/4)

krok68 [10]

$Tan(x-\frac{\pi }{4}) = \frac{-17}{31}$

<u>Step-by-step explanation:</u>

Here we have , sin ⁡x=7/25( given sin x = 725 which is not possible ) , $0$ . Let's find tan (x - pi/4):

⇒ $Tanx = \frac{sinx}{cosx}$

⇒ $Tanx = \frac{sinx}{\sqrt{1-(sinx)^{2}}}$

⇒ $Tanx = \frac{\frac{7}{25}}{\sqrt{1-(\frac{7}{25})^{2}}}$

⇒ $Tanx = {\frac{7}{25}}{\sqrt{(\frac{625}{625-49})^{}}}$

⇒ $Tanx = {\frac{7}{25}}(\frac{25}{24} )$

⇒ $Tanx = {\frac{7}{24}}$

Now , $Tan(x-\frac{\pi }{4}) = \frac{Tanx - Tan(\frac{\pi }{4} )}{1+ Tanx(Tan(\frac{\pi }{4} )}$

⇒ $Tan(x-\frac{\pi }{4}) = \frac{Tanx -1}{1+ Tanx(1)}$

⇒ $Tan(x-\frac{\pi }{4}) = \frac{\frac{7}{24} -1}{1+\frac{7}{24} }$

⇒ $Tan(x-\frac{\pi }{4}) = \frac{\frac{7-24}{24} }{\frac{7+24}{24} }$

⇒ $Tan(x-\frac{\pi }{4}) = \frac{-17}{24} (\frac{24}{31} )$

⇒ $Tan(x-\frac{\pi }{4}) = \frac{-17}{31}$

7 0

3 years ago

What is the area of this shape?

goldenfox [79]

I got 132 for the area of this shape

6 0

3 years ago

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