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lana66690 [7]
3 years ago
5

The ordered pair (a,b) satisfies the inequality y>x+3. Which statement is true ?

Mathematics
1 answer:
nordsb [41]3 years ago
3 0

Answer:The correct option is(C). b is greater than 3.

Step-by-step explanation:

 Given that the ordered pair (a, b) satisfied the following inequality :

We are to select the TRUE statement from the given options.

Since the ordered pair (a, b) satisfies inequality (i), so we have

Therefore, b is greater than 3.

Thus, the correct option is (C). b is greater than 3.

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xenn [34]

Answer:

B=5,0.

Step-by-step explanation:

It's not that hard. It's just that it might throw you off a bit with the R and the other things. Functions if you don't remember are where the coordinates don't have the same x and y. So since the y1 and y2 are different then to make it not a function, you have to make x1 and x2. To make it that b would have to be = to 5 and 0.

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3 years ago
Rosa ran the hurdles in 58.32 seconds. maya ran the hurdles in 54.078 seconds. what is the difference between the two times? 3.9
satela [25.4K]

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5 0
1 year ago
(3x + 3x²) + (2x² + 2x)?​
Galina-37 [17]

Answer:

=5x2+5x

Step-by-step explanation:

Let's simplify step-by-step.

3x+3x2+2x2+2x

8 0
3 years ago
In parallelogram DEFG, DH = x + 5, HF = 2y, GH = 3x – 1, and HE = 5y + 4. Find the values of x and y.
earnstyle [38]
The diagonals of a parallelogram bisect each other.

DH = HF and GH = HE

x + 5 = 2y
3x - 1 = 5y + 4

Solve the first equation for x.
x = 2y - 5

Now substitute 2y - 5 for x in the second equation.

3(2y - 5) - 1 = 5y + 4

6y - 15 - 1 = 5y + 4

6y - 16 = 5y + 4

y = 20

Now substitute 20 for y in the first original equation.

x + 5 = 2y

x + 5 = 2(20)

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Answer: x = 35 and y = 20
4 0
3 years ago
Read 2 more answers
Quadrant:
NISA [10]

<h2>✒️Area Between Curves</h2>

\small\begin{array}{ |c|c} \hline \bold{Area\ Between\ Curves} \\ \\ \textsf{Solving for the intersection of }\rm y = x^2 + 2\textsf{ and }\\ \rm y = 4, \\ \\ \qquad \begin{aligned} \rm y_1 &=\rm y_2 \\ \rm x^2 + 2 &=\rm 4 \\ \rm x^2 &= \rm 2 \\ \rm x &=\rm \pm \sqrt{2} \end{aligned} \\ \\ \textsf{We only need the first quadrant area bounded} \\ \textsf{by the given curves so the integral for the area} \\ \textsf{would then be} \\ \\ \boldsymbol{\displaystyle \rm A = \int_{\ a}^{\ b} {\left( \begin{array}{c}\text{upper} \\ \text{function}\end{array} \right) - \left( \begin{array}{c} \text{lower} \\ \text{function} \end{array} \right)\ dx}} \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} \Big[4 - (x^2 + 2)\Big]\ dx \\ \\ \displaystyle \rm A = \int_{0}^{\sqrt{2}} (2 - x^2)\ dx \\ \\ \rm A = \left[2x - \dfrac{x^3}{3}\right]_{0}^{\sqrt{2}} \\ \\ \rm A = 2\sqrt{2} - \dfrac{\big(\sqrt{2}\big)^3}{3} \\ \\ \rm A = 2\sqrt{2} - \dfrac{2\sqrt{2}}{3} \\ \\\red{\boxed{\begin{array}{c} \rm A = \dfrac{4\sqrt{2}}{3}\textsf{ sq. units} \\ \textsf{or} \\ \rm A \approx 1.8856\textsf{ sq. units} \end{array}}} \\\\\hline\end{array}

#CarryOnLearning

#BrainlyForTrees

\qquad\qquad\qquad\qquad\qquad\qquad\tt{Monday\:at \: 04-04-2022} \\ \qquad\qquad\qquad\qquad\qquad\qquad\tt{12:10 \: pm}

5 0
2 years ago
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