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ZanzabumX [31]
2 years ago
8

The graph of the function has slope of 1 and y-intercept of -6.​

Mathematics
1 answer:
True [87]2 years ago
6 0

Answer:

= -5

Step-by-step explanation:

1-6=5

1-6

{Subtract the numbers:} 1-6=-5

= -5

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3/v=2/4? i need help plz
aalyn [17]

Answer:

v = 6

Step-by-step explanation:

3/v = 2/4

We can use cross products to solve

2v = 3*4

2v = 12

Divide each side by 2

2v/2 = 12/2

v =6

6 0
2 years ago
Read 2 more answers
Find the approximate length of segment AC. Math item stem image CLEAR CHECK 5.74 units 8.06 units 11 units 65 units
laiz [17]

Answer:

         \large\boxed{\large\boxed{8.06units}}

Explanation:

The figure is a right triangle with:

  • hypotenuse = segment AC
  • vertical leg = 7 units
  • horizontal leg = 4 units

Hence, you must use Pythagora's theorem, which is valid for all right triangles:

             hypotenuse^2=(leg_1)^2+(leg_2)^2

Substituting the data:

        hypotenuse^2=(7unit)^2+(4unit)^2=49unit^2+16unit^2=65unit^2

Square root both sides:

         hypotenuse=\sqrt{65unit^2}=8.06unit

8 0
3 years ago
What is the common ratio of the geometric sequence below?<br><br> 625, 125, 25, 5, 1, ...
larisa [96]

Answer: 1/5 or 0.2

Step-by-step explanation: each number is just dividing by 5

like 625/5 = 125

125/5 = 25

25/5 =5

5/5 = 1

similarly 1/5 = 0.2 or 20%

4 1
3 years ago
harry is trying tq solve the equation y=2x^2 -x-6 using the quadratic formula he has made an error in one of the steps below
Sergeeva-Olga [200]

Answer:

Hello :

y = 2x² - x -6

y = 2 ( x² - 1/2 x - 3)

y = 2 (x² -1/2 x + 1/16 -1/16 -3)

y = 2 (( x - 1/4 )² - 49/16).....  the quadratic formula

y = 0 :  ( x - 1/4 )² - 49/16 = 0

( x - 1/4 )² - (7/4)² = 0

( x -1/4-7/4)( x -1/4+7/4)=0

(x -2)(x + 3/2) = 0

x - 2 = 0 or x + 3/2 =0

x = 2 or x = - 3/2



3 0
3 years ago
Find the derivative
zalisa [80]

First use the chain rule; take y=\dfrac{x+5}{x^2+3}. Then

\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dy}\cdot\dfrac{\mathrm dy}{\mathrm dx}

By the power rule,

f(x)=y^2\implies\dfrac{\mathrm df}{\mathrm dy}=2y=\dfrac{2(x+5)}{x^2+3}

By the quotient rule,

y=\dfrac{x+5}{x^2+3}\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{(x^2+3)\frac{\mathrm d(x+5)}{\mathrm dx}-(x+5)\frac{\mathrm d(x^2+3)}{\mathrm dx}}{(x^2+3)^2}

\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{(x^2+3)-(x+5)(2x)}{(x^2+3)^2}

\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3-10x-x^2}{(x^2+3)^2}

So

\dfrac{\mathrm df}{\mathrm dx}=\dfrac{2(x+5)}{x^2+3}\cdot\dfrac{3-10x-x^2}{(x^2+3)^2}

\implies\dfrac{\mathrm df}{\mathrm dx}=\dfrac{2(x+5)(3-10x-x^2)}{(x^2+3)^3}

6 0
3 years ago
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