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2 years ago

(25a + 9d) - (10a - 5d)

Mathematics

2 answers:

Anton [14]2 years ago

4 0

Answer:

15a+14d

Step-by-step explanation:

I just combined like terms

stealth61 [152]2 years ago

4 0

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Convert equation from slope intercept form to standard form y=-7/3x+1

Andrews [41]

Answer:

7x + 3y = 3

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2 years ago

Please Help!

alisha [4.7K]

Diagonal of the parallelogram divides the parallelogram in to two equal areas.

So area of parallelogram = 2(area of triangle)

According to the given diagram,

AB= 8, AD = 5 and BD = 11

So according to the Heron's formula,

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}\\\\where,\\ s =\frac{a + b + c }{2}$

and a, b and c are the three sides of the triangle

Area of triangle ABD = $s=\frac{8 + 5 + 11}{2} \\\\s = \frac{24}{2}\\\\s = 12Area of triangle ABD = \sqrt{12(12 - 8) (12 - 5) (12 - 11)}\\\\Area of triangle ABD = \sqrt{12(4) (7) (1)}\\\\Area of triangle ABD = \sqrt{336}\\\\Area of triangle ABD = 18.33$

So, area of parallelogram ABCD = 2(area of triangle ABD)

area of parallelogram ABCD = 2 (18.33)

area of parallelogram ABCD = 36.66

area of parallelogram ABCD = 36.7 sq. units

6 0

2 years ago

Read 2 more answers

Which is true regarding the system of equations?<br><br> x minus 4 y = 1. 5 x minus 20 y = 4.

goldfiish [28.3K]

Answer:

No solution

Step-by-step explanation:

x - 4y = 1 --> (1)

5x - 20y = 4 --> (2)

y = (¼)x - ¼ --> (1)

y = (¼)x - ⅕ --> (2)

Since these are 2 parallel lines with different y-intercepts, they will never meet

3 0

2 years ago

Read 2 more answers

A trough of water is 20 meters in length and its ends are in the shape of an isosceles triangle whose width is 7 meters and heig

Vaselesa [24]

Answer:

a) Depth changing rate of change is $0.24m/min$ , When the water is 6 meters deep

b) The width of the top of the water is changing at a rate of $0.17m/min$ , When the water is 6 meters deep

Step-by-step explanation:

As we can see in the attachment part II, there are similar triangles, so we have the following relation between them $\frac{3.5}{10} =\frac{a}{h}$ , then $a=0.35h$ .

a) As we have that volume is $V=\frac{1}{2} 2ahL=ahL$ , then $V=(0.35h^{2})L$ , so we can derivate it $\frac{dV}{dt}=2(0.35h)L\frac{dh}{dt}$ due to the chain rule, then we clean this expression for $\frac{dh}{dt}=\frac{1}{0.7hL}\frac{dV}{dt}$ and compute with the knowns $\frac{dh}{dt}=\frac{1}{0.7(6m)(20m)}2m^{3}/min=0.24m/min$ , is the depth changing rate of change when the water is 6 meters deep.

b) As the width of the top is $2a=0.7h$ , we can derivate it and obtain $\frac{da}{dt}=0.7\frac{dh}{dt} =0.7*0.24m/min=0.17m/min$ The width of the top of the water is changing, When the water is 6 meters deep at this rate

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2 years ago

Which function is the inverse of f(x) = –5x – 4?

makvit [3.9K]

Y = -5x - 4
switch x and y and solve for y
x = -5y - 4
x+4 = -5y
(x+4)/ -5 = y^-1 ( the inverse)

4 0

3 years ago

(25a + 9d) - (10a - 5d)​

(25a + 9d) - (10a - 5d)