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Nikolay [14]
3 years ago
14

GUYS I NEED HELP ITS URGENT! WILL MARK BRAINLIEST!

Mathematics
1 answer:
Ad libitum [116K]3 years ago
7 0

Answer:

k = 13 The smallest zero or root is x = -10

===============================

Work Shown:

note: you can write "x^2" to mean "x squared"

f(x) = x^2+3x-10

f(x+5) = (x+5)^2+3(x+5)-10 ... replace every x with x+5

f(x+5) = (x^2+10x+25)+3(x+5)-10

f(x+5) = x^2+10x+25+3x+15-10

f(x+5) = x^2+13x+30

Compare this with x^2+kx+30 and we see that k = 13

Factor and solve the equation below

x^2+13x+30 = 0

(x+10)(x+3) = 0

x+10 = 0 or x+3 = 0

x = -10 or x = -3

The smallest zero is x = -10 as its the left-most value on a number line.

Hope this was Right

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What is the slope / rate of change for 2, 5 and -3, 7​
Nady [450]

Answer:

\displaystyle m=\frac{-2}{5}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS<u> </u>

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

  • Coordinates (x, y)
  • Slope Formula: \displaystyle m=\frac{y_2-y_1}{x_2-x_1}

Step-by-step explanation:

<u>Step 1: Define</u>

Point (2, 5)

Point (-3, 7)

<u>Step 2: Identify</u>

x₁ = 2, y₁ = 5

x₂ = -3, y₂ = 7

<u>Step 3: Find slope </u><em><u>m</u></em>

Simply plug in the 2 coordinates into the slope formula to find slope<em> m</em>

  1. Substitute in points [Slope Formula]:                                                             \displaystyle m=\frac{7-5}{-3-2}
  2. [Slope] [Fraction] Subtract:                                                                             \displaystyle m=\frac{2}{-5}
  3. [Slope] [Fraction] Rewrite:                                                                              \displaystyle m=\frac{-2}{5}
4 0
3 years ago
Kevin counts 75 chicks in the hen house. He also sees more chicks outside the hen house. How can Kevin count to find how many ch
stellarik [79]

What we know:

we know that there are <u>75</u> chicks in the hen house, we also know that in all there are 108 chicks

What we need to find:

we need to find out how many chicks were outside

Answer equation:

108 - 75 = 33

8 0
3 years ago
A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,
guajiro [1.7K]

We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation A=45e^{-0.05t}.

To find the time it will take for the area of the glacier to decrease to 15 square kilo-meters, we will equate A=15 and solve for t as:

15=45e^{-0.05t}

\frac{15}{45}=\frac{45e^{-0.05t}}{45}

\frac{1}{3}=e^{-0.05t}

Now we will switch sides:

e^{-0.05t}=\frac{1}{3}

Let us take natural log on both sides of equation.

\text{ln}(e^{-0.05t})=\text{ln}(\frac{1}{3})

Using natural log property \text{ln}(a^b)=b\cdot \text{ln}(a), we will get:

-0.05t\cdot \text{ln}(e)=\text{ln}(\frac{1}{3})

-0.05t\cdot (1)=\text{ln}(\frac{1}{3})

-0.05t=\text{ln}(\frac{1}{3})

t=\frac{\text{ln}(\frac{1}{3})}{-0.05}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-0.05\cdot 100}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-5}

t=-\text{ln}(\frac{1}{3})\cdot 20

t=-(\text{ln}(1)-\text{ln}(3))\cdot 20

t=-(0-\text{ln}(3))\cdot 20

t=20\text{ln}(3)

Therefore, it will take 20\text{ln}(3) years for area of the glacier to decrease to 15 square kilo-meters.

6 0
3 years ago
What is the answer pls I will mark as brainless
frosja888 [35]

Answer:

-10

Step-by-step explanation:

24+8-(7x6)

3 0
3 years ago
the measurements of a photo and it's frame are shown in the diagram. Write a polynomial that represents the width of the photo.
suter [353]

Answer:

The width of the photo is 4w^2+6w+4.

Step-by-step explanation:

From the given figure it is notices that the total width of the frame is

6w^2+8

The photo is covered by a frame border and the width of the border is

w^2-3w+2

To find the width of the photo we have to subtract the width of upper frame border and lower frame border from the total width of frame.

Width of the photo is

\text{Width of the photo}=\text{Width of the frame}-2(\text{Width of the frame border})

\text{Width of the photo}=6w^2+8-2(w^2-3w+2)

\text{Width of the photo}=6w^2+8-2w^2+6w-4

\text{Width of the photo}=4w^2+6w+4

Therefore the width of the photo is 4w^2+6w+4.

4 0
3 years ago
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