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3 years ago

Question:

Mathematics

1 answer:

mylen [45]3 years ago

5 0

At least up to 100 miled

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Simplify to create a equivalent expression.

Ratling [72]

Answer:

$\boxed{\bold{66k+9}}$

Explanation:

Expand $\bold{5\left(10k+1\right): \ 50k+5}$

= $\bold{50k+5+2\left(2+8k\right)}$

Expand $\bold{2\left(2+8k\right): \ 4+16k}$

= $\bold{50k+5+4+16k}$

Simplify $\bold{50k+5+4+16k: \ 66k+9}$

= $\bold{66k+9}$

Mordancy.

7 0

3 years ago

Read 2 more answers

Find the midpoint of the line segment joining the points R(-5,5) and S(4.6).

tigry1 [53]

Hey there! I'm happy to help!

To find the x value of the midpoint, you add the x values and divide by 2.

-5+4=-1

-1/2=-1/2

For the y value, you add the y values and divide by 2.

5+6=11

11/2=5 1/2

So, the midpoint is (-1/2, 5 1/2).

Have a wonderful day! :D

6 0

3 years ago

Maria mixes 1/4 cup of white sugar and 2/3 cup of brown sugar, to make a topping for some muffins. She uses 1/12 cup of the mix

Sergeu [11.5K]

Maria can top 11 muffins.

7 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 $x\in A\cap B$ .