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3 years ago

When finding the equation of a linear function in a table how do I find y when x is 0?

Mathematics

1 answer:

sesenic [268]3 years ago

3 0

U need to find the equation of the line....and once u do that, sub in 0 for x and solve for y

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In the given figure △ABC ≅△DEC. Which of the following relations can be proven using CPCTC ?

Serhud [2]

Option B:

$\overline{A B}=\overline{D E}$

Solution:

In the given figure $\triangle A B C \cong \triangle D E C$ .

If two triangles are similar, then their corresponding sides and angles are equal.

By CPCTC, in $\triangle A B C \ \text{and}\ \triangle D E C$ ,

$\overline{AB }=\overline{DE}$ – – – – (1)

$\overline{B C}=\overline{EC}$ – – – – (2)

$\overline{ CA}=\overline{CD}$ – – – – (3)

$\angle ACB=\angle DCE$ – – – – (4)

$\angle ABC=\angle DEC$ – – – – (5)

$\angle BAC=\angle EDC$ – – – – (6)

Option A: $\overline{B C}=\overline{D C}$

By CPCTC proved in equation (2) $\overline{B C}=\overline{EC}$ .

Therefore $\overline{B C}\neq \overline{D C}$ . Option A is false.

Option B: $\overline{A B}=\overline{D E}$

By CPCTC proved in equation (1) $\overline{AB }=\overline{DE}$ .

Therefore Option B is true.

Option C: $\angle A C B=\angle D E C$

By CPCTC proved in equation (4) $\angle ACB=\angle DCE$ .

Therefore $\angle A C B\neq \angle D E C$ . Option C is false.

Option D: $\angle A B C=\angle E D C$

By CPCTC proved in equation (5) $\angle ABC=\angle DEC$ .

Therefore $\angle A B C\neq \angle E D C$ . Option D is false.

Hence Option B is the correct answer.

$\Rightarrow\overline{A B}=\overline{D E}$

5 0

3 years ago

A paper clip is dropped from the top of a 144‐ft tower, with an initial velocity of 16 ft/sec. Its position function is s(t) = −

hoa [83]

To calculate the velocity, we use the given expression above which is s(t) = −16t^2 + 144. First, we calculate the time it takes to reach the ground. Then, differentiate the expression and substitute time to the differentiated expression.

s(t) = −16t^2 + 144
0 = -16t^2 + 144
t = 3

s'(t) = v = -32t
v = -32(3)
v = -96

Note: negative sign signifies that the object is going down

5 0

2 years ago

Read 2 more answers

Help algebra 2 please

kati45 [8]

Answer:

Step-by-step explanation:

(f*g)(x) = (-5x² + 2x + 7) (x +1)

= x* (-5x² + 2x + 7) + 1*(-5x² + 2x + 7)

= x*(-5x²) + x*2x + x*7 - 5x² + 2x + 7

= -5x³ + 2x² + 7x - 5x² + 2x + 7

= - 5x³ + 2x² -5x² + 7x + 2x +7 {Combine like terms}

= -5x³ - 3x² + 9x + 7

4) (f*g)(x) = (x² + 2x + 4)(x - 2)

= x*(x² + 2x + 4) - 2*(x² + 2x + 4)

= x*x² + x*2x + x*4 - 2*x² - 2*2x -2* 4

= x³ + 2x² + 4x -2x² -4x - 8

= x³ - 8

$2) Speed=\dfrac{distance}{time}\\\\\\= \dfrac{d(h)}{t(h)}\\\\= \dfrac{2\sqrt{h}}{3\sqrt{4h}}=\dfrac{2*\sqrt{h}}{3*2\sqrt{h}}\\\\\\=\dfrac{1}{3}$

$1)d(h) = \sqrt{16h^{4}}=\sqrt{2*2*2*2*h*h*h*h}=2*2*h*h=4h^{2}\\\\\\t(h) = 3h^{4}-3h^{3}-5h^{2}= h^{2}(3h^{2}-3h - 5))\\\\Speed=\dfrac{4h^{2}}{h^{2}(3h^{2}-3h-5)}\\\\=\dfrac{4}{3h^{2}-3h-5}$