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VLD [36.1K]
2 years ago
12

(3, -3); m = -2 In standard form

Mathematics
1 answer:
xz_007 [3.2K]2 years ago
6 0
In y=mx +b?

If so then -3=-2(3)+b
B=3
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A car can be rented for $75 per week plus $0.35 per mile. How many miles can be driven if you have at most $110 to spend for wee
sergey [27]

After you subtract the fixed cost of $75 from your budget of $110, you have $35 to spend on miles. At $0.35 per mile, you can afford 100 of them.

You can drive up to 100 miles for $110 or less.

8 0
3 years ago
Theresa Mitsoto invested $3,500 in a 3-year CD that paid 6.5% annual interest. She cashed out the CD at the end of two years wit
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Answer:

$56.875

Step-by-step explanation:

Given that :

Amount invested = principal, p = 3500

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7 0
2 years ago
The function f(x) = –x2 – 4x + 5 is shown on the graph. On a coordinate plane, a parabola opens down. It goes through (negative
sertanlavr [38]

The true statement about the function f(x) = -x² - 4x + 5 is that:

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<h3 /><h3>What is the domain and range for the function of y = f(x)?</h3>

The domain of a function is the set of given values of input for which the function is valid and true.

The range is the dependent variable of a given set of values for which the function is defined.

  • The domain of the function: f(x) = -x² - 4x + 5 are all real number from -∞ to +∞

For a parabola ax² + bx + c  with the vertex \mathbf{(x_v,y_v)}

  • If a < 0, then the range is f(x) ≤ \mathbf{y_v}
  • If a > 0, then the range f(x) ≥  \mathbf{y_v}
  • Here; a = -1,

The vertex for an up-down facing parabola for a function y = ax² + bx + c is:

\mathbf{x_v = -\dfrac{b}{2a}}

Thus,

  • vertex \mathbf{(x_v,y_v)} = (-2, 9)

Range: f(x) ≤ 9

Therefore, we can conclude that the range of the function is all real numbers less than or equal to 9.

Learn more about the domain and range of a function here:

brainly.com/question/26098895

#SPJ1

5 0
2 years ago
Write the ordered pair for a point that is 9 units to the left of the origin and lies on the x-axis.
natulia [17]
(((((((((-9,0)))))))))
8 0
3 years ago
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