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salantis [7]
3 years ago
12

How do I solve (4x+1)(x-5)=-43x-41

Mathematics
1 answer:
AfilCa [17]3 years ago
8 0
Remember you can do anything to an equation as long as you do it to both sides
expand by foiling or distributing


4x²-19x-5=-43x-41
add 43x to both sides and add 41 to both sides

4x²+24x+36=0
factor
facotr out of 4
4(x²+6x+9)=0
factor inside
what 2 numbers multiply to get 9 and add to get 6?
3 and 3
4(x+3)(x+3)=0
set to 0

x+3=0
x=-3

x=-3 is answer
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Here are two shapes with the same area. Work out the perimiter of the rectangle.
Lilit [14]
So, we know the perimeter is 24 cm. and the top part is the left part, (x) + 2, but remember, we need to have it be a number that, multiplied by 2, is 24. So, using this formula, the length, (top / bottom) is 7, and x = 5. Because 5 + 5 = 10, and when length is +2, it adds up to 14. 10 + 14 = 24. So, length is 7, width, (left, right) is 5.
3 0
3 years ago
19. Ms. Porter had eight parties at her house last year. The number of guests at each party
KatRina [158]

Answer:

17

Step-by-step explanation:

Order the numbers from Smallest to Largest...

9, 10, 14, 15, 19, 21, 21, 27

Then the middle number is your answer.

Since there are 2 Middle Numbers, you add them together, and divide by 2.

Which would be, 17.

6 0
3 years ago
Read 2 more answers
In triangle abc, cos A= -0.6. Find A and Tan A
Tresset [83]

Answer:

Part 1) A=126.87^o

Part 2) tan(A)=-\frac{4}{3}

Step-by-step explanation:

we have

cos(A)=-0.6

The cos(A) is negative, that means that the angle A in the triangle ABC is an obtuse angle and the value of the sin(A) is positive

The angle A lie on the II Quadrant

step 1

Find the measure of angle A

cos(A)=-0.6

using a calculator

A=cos^{-1}(-0.6)=126.87^o

step 2

Find the sin(A)

we know that

sin^2(A)+cos^2(A)=1

substitute the value of cos(A)

sin^2(A)+(-0.6)^2=1

sin^2(A)=1-0.36

sin^2(A)=0.64

sin(A)=0.8

step 3

Find tan(A)

we know that

tan(A)=\frac{sin(A)}{cos(A)}

substitute the values

tan(A)=\frac{0.8}{-0.6}

Simplify

tan(A)=-\frac{4}{3}

5 0
3 years ago
For brainiest:):):):):):):)
exis [7]
I used 3.14 for pi to solve easier.

6. 6.28

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6 0
3 years ago
A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height
Crank

Answer:

a) h = 0.1: \bar v = -11\,\frac{ft}{s}, h = 0.01: \bar v = -10.1\,\frac{ft}{s}, h = 0.001: \bar v = -10\,\frac{ft}{s}, b) The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

Step-by-step explanation:

a) We know that y = 30\cdot t -10\cdot t^{2} describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity (\bar v), measured in feet per second, can be done by means of the following definition:

\bar v = \frac{y(2+h)-y(2)}{h}

Where:

y(2) - Position of the ball evaluated at t = 2\,s, measured in feet.

y(2+h) - Position of the ball evaluated at t =(2+h)\,s, measured in feet.

h - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

h = 0.1\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}

y(2.1) = 18.9\,ft

\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}

\bar v = -11\,\frac{ft}{s}

h = 0.01\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}

y(2.01) = 19.899\,ft

\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}

\bar v = -10.1\,\frac{ft}{s}

h = 0.001\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}

y(2.001) = 19.99\,ft

\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}

\bar v = -10\,\frac{ft}{s}

b) The instantaneous velocity when t = 2\,s can be obtained by using the following limit:

v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}

v(t) =  \lim_{h \to 0} 30-20\cdot t-10\cdot h

v(t) = 30\cdot  \lim_{h \to 0} 1 - 20\cdot t \cdot  \lim_{h \to 0} 1 - 10\cdot  \lim_{h \to 0} h

v(t) = 30-20\cdot t

And we finally evaluate the instantaneous velocity at t = 2\,s:

v(2) = 30-20\cdot (2)

v(2) = -10\,\frac{ft}{s}

The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

8 0
3 years ago
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