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2 years ago

Find AB, BC, and AC

Mathematics

1 answer:

sp2606 [1]2 years ago

4 0

This algebra one or two please mark me brainlist I need 1

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Pls pls help it is due todayyy

Pani-rosa [81]

Answer:

I did this today and its H

Step-by-step explanation:

Please lmk if im wrong, but im sure its right but js ;)

6 0

2 years ago

Y = 2x2 - 3x - 20?

nikitadnepr [17]

2x² - 3x - 20

40
2 20
4 10
5 8

2x² - 8x 5x - 20
2x(x - 4) 5(x - 4)
(2x + 5) (x - 4)
2x + 5 = 0 x - 4 = 0
2x = -5 x = 4
x = -5/2
C) x = -2.5 and 4

4 0

2 years ago

Which expressions are equivalent Log 2- log 8

Aleksandr [31]

Answer:

log(.25)

Step-by-step explanation:

$log(a)-log(b)=log(\frac{a}{b})\\log(2)-log(8)=log(\frac{2}{8})\\log(\frac{1}{4})$

8 0

2 years ago

A penny is dropped from the empire state building which is 1250 feet tall if the penny's pathway can be modeled by the equation

yaroslaw [1]

Answer:

It would take approximately 8.82 sec for a penny to strike the person.

Step-by-step explanation:

Given:

Height of the building = 1250 feet

Height of the person = 6 feet.

Modeled equation is $h=-16t^2+1250$

We need to find the time required 't' for the penny to strike the person.

Solution:

to find the time required 't' for the penny to strike the person we will substitute;

h ⇒ height of the person.

So we get;

$6=-16t^2+1250$

Now Combining the like terms we get;

$16t^2=1250-6\\\\16t^2=1244$

Dividing both side by 16 using division property we get;

$\frac{16t^2}{16} =\frac{1244}{6}\\\\t^2 = 77.75$

Now taking square root on both side we get;

$\sqrt{t^2}= \sqrt{77.75}\\\\t \approx 8.82 secs$

Hence It would take approximately 8.82 sec for a penny to strike the person.

8 0

2 years ago

Y=−3(2.5)x

Hoochie [10]

Answer:

1. The equation represent an exponential decay

2. The rate of the exponential decay is -3×2.5ˣ·㏑(2.5)

Step-by-step explanation:

When a function a(t) = a₀(1 + r)ˣ has exponential growth, the logarithm of x grows with time such that;

log a(t) = log(a₀) + x·log(1 + r)

Hence in the equation -3 ≡ a₀, (1 + r) ≡ 2.5 and y ≡ a(t). Plugging in the values in the above equation for the condition of an exponential growth, we have;

log y = log(-3) + x·log(2.5)

Therefore, since log(-3) is complex, the equation does not represent an exponential growth hence the equation represents an exponential decay.

The rate of the exponential decay is given by the following equation;

$\frac{dy}{dx} =\frac{d(-3(2.5)^x)}{dx} = -\frac{d(3\cdot e^{x\cdot ln(2.5)})}{dx} = -3 \times 2.5^x\times ln(2.5)$

Hence the rate of exponential decay is -3×2.5ˣ × ㏑(2.5)

5 0

3 years ago