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ANTONII [103]
3 years ago
9

Find the area of a rectangle with length (2x + 3) and width (x - 4).

Mathematics
1 answer:
slava [35]3 years ago
5 0

Answer:

2x^2-5x-12

Step-by-step explanation:

a rectangle has a length of 2x+3 and a width of x-4 find the area

------------

Area = length * width

-------------------------

A = (2x+3)(x-4)

---

A = 2x(x-4)+3(x-4)

 

---

A = 2x^2-8x+3x-12

---

A = 2x^2-5x-12

-------------------

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7×300=7× blank hundreds
VladimirAG [237]
7x300=7x 3 hundres because in 300 hundreds there is 3 hundreds.I know that didnt make sense.so basically just look at the front number number its 3 so 3 hundreds but if your working with thousands and the front number is 5 its 5 thousands.I hope this was helpful.
6 0
3 years ago
B) Here are the first four terms of a quadratic sequence.
Nookie1986 [14]

Find answer is the given attachment

5 0
3 years ago
Stewart said that 30-10p does not equal 5(2p-6) for any value of p. Do you agree with Stewart? Justify Stewart's answer, or expl
charle [14.2K]
Ya he is correct !

for any value of P, both can never be equal !
as, 5(2p-6) = 10p-30 .....so even if we put any value , negative or positive, (-30) can not be changed anyway !

thus 30-10p can never be equal to 5(2p-6) !!
5 0
3 years ago
What is simplest form of sqrt 2 + sqrt 5 / sqrt 2 - sqrt -5? √2 + √5 / √2 - √-5
Leni [432]

Answer:

\frac{-7-2\sqrt{10}}{3}

Step-by-step explanation:

we have

\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}-\sqrt{5}}

Simplify

Multiply the expression by  \frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}}

(\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}-\sqrt{5}})(\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}})

Apply difference of squares in the denominator

\frac{(\sqrt{2}+\sqrt{5})^2}{(\sqrt{2})^2-(\sqrt{5})^2}

\frac{2+2\sqrt{10}+5}{2-5}

\frac{7+2\sqrt{10}}{-3}

\frac{-7-2\sqrt{10}}{3}

6 0
3 years ago
Find the distance between the two points S(-5, -2) and T(-3, 4).
Dafna1 [17]

Step-by-step explanation:

Hey there!

The points are; S(-5,-2) and T(-3,4).

Using distance formula to find the distance between two points.

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

Put all values.

d =  \sqrt{  {( - 3 + 5)}^{2}  +  {(4 + 2)}^{2} }

Simplify it.

d =  \sqrt{ {(2)}^{2} +  {(6)}^{2}  }

d =  \sqrt{4 + 36}

d =  \sqrt{40}

Therefore the distance between two points is root 40 units Or 6.324 units.

Or, you can write as 6 units after rounding off.

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

5 0
3 years ago
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