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3 years ago

Will AWARD BRAINIEST!!!!! 20 POINTS

Mathematics

1 answer:

Vesna [10]3 years ago

5 0

Answer:

Step-by-step explanation:

The 3 by 2 rectangle is 6 so 15-6=9

3 x (x-2) x 0.5 = 9

3 x (6-2=4) 4 = 12 / 2 = 6

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Is h(x)=3x^3+2x^2+5 even, odd, or neither?

pochemuha

Answer:

Neither even or odd

Step-by-step explanation:

4 0

3 years ago

CAN SOMEONE HELP ME IN THIS QUESTION?

svetoff [14.1K]

I
I'm pretty sure it is B.

8 0

3 years ago

The equation of a circle is x2 + y2 + Cx + Dy + E = 0. If the radius of the circle is decreased without changing the coordinates

valkas [14]

Answer:

Option E

Step-by-step explanation:

The standard equation of circle is:- (x-h)² + (y-k)² = r²

where (h,k) is center point and r is radius

If radius r is decreased then also (h,k) remains same, only r² increases

Equation given in question is x² + y² + Cx + Dx + E=0

So, C and D will remain same but E will increase

4 0

3 years ago

Nine more than half of a number is 21

pogonyaev

I assume you're looking for the number:
0.5n + 9 = 21
- 9
0.5n = 12
÷ 0.5
n = 24
So the beginning number is 24, hope this helps!

3 0

3 years ago

Let A = { x ∈ R : x 2 − 5 x + 4 ≤ 0 } , B = (3 , 5), and C = (3 , 4]. Show that A ∩ B = C . (Recall, you must show two separate

kolbaska11 [484]

Answer:

You can proceed as follows:

Step-by-step explanation:

First solve the quadratic inequality $x^{2}-5x+4\leq 0$ . To do that, factorize, then we have that $(x-4)(x-1)\leq 0$ . This implies that

$x-1\leq 0\, \text{and}\, x-4\geq 0$

$x-1 \geq 0\, \text{and}\, x-4\leq 0$

In the first case the solution is the empty set $\emptyset$ . In the second case the solution is the interval $1\leq x \leq 4$ . Now we have that

$A=[1,4]$

$B=(3,5)$

$C=(3,4]$ .

To show that $A\cap B\subseteq C$ consider $x\in A\cap B$ . Then $1\leq x \leq 4\, \text{and}\, 1$ , this implies that $3$ , then $x\in C$ . Now, to show that $C\subseteq A\cap B$ consider $x\in C$ , then $3$ , then $1\leq x \leq 4\, \text{and}\, 3$ , then $x\in [1,4] \, \text{and}\, x\in (3,5)$ , this implies that $x\in A\cap B$ .

Observe the image below.

7 0

3 years ago