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Iteru [2.4K]
2 years ago
10

Twice a number decreased by seven

Mathematics
1 answer:
11Alexandr11 [23.1K]2 years ago
5 0
The equation would be 2n-7
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I need answers anyone help
GalinKa [24]
First one single solution second no solution 3rd one single forth one infinity and last 2 are single and infinity
7 0
3 years ago
Please choose answer choice a,b,c or d
dmitriy555 [2]

Answer:

A. -4

Step-by-step explanation:

Slope is rise over run, or y2-y1 / x2-x1.

\frac{y2-y1}{x2-x1} = \frac{-4-8}{2--1} =\frac{-12}{3} = -4

<u>The slope is A. -4.</u>

Let's see--does that make sense? A slope of four would be pretty steep, and that matches the graph. It also should be negative because the line goes down and to the right, not up and to the right, so -4 makes sense!

4 0
2 years ago
Evaluate the expression
lora16 [44]
The correct answer is B=4
5 0
2 years ago
Write the slope intercept form equation of the line given in graph
Romashka [77]

Answer:

y=-1

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Find the derivative.
Aleksandr [31]

Answer:

Using either method, we obtain:  t^\frac{3}{8}

Step-by-step explanation:

a) By evaluating the integral:

 \frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du

The integral itself can be evaluated by writing the root and exponent of the variable u as:   \sqrt[8]{u^3} =u^{\frac{3}{8}

Then, an antiderivative of this is: \frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}

which evaluated between the limits of integration gives:

\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}

and now the derivative of this expression with respect to "t" is:

\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

g(x)=\int\limits^x_a {f(t)} \, dt

is continuous on [a,b], differentiable on (a,b) and  g'(x)=f(x)

Since this this function u^{\frac{3}{8} is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}

5 0
3 years ago
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