What is the answer for Chris' age and Mary's age?

Mathematics

2 answers:

kati45 [8]3 years ago

8 0

M=mary's age
c=chris's age

m is 4 younger than 3 times chris
m=-4+3c
m=3c-4

mary's age 3 years ago (m-3) is equal to chris age in 9 years (c+9), find c and m

m-3=c+9

we got
m=3c-4
sub 3c-4 for m in other eqution
3c-4-3=c+9
3c-7=c+9
minus c both sides
2c-7=9
add 7 to both sides
2c=16
divide both sides by 2
c=8

sub back
m=3c-4
m=3(8)-4
m=24-4
m=20

mary=20
chris=8

Allushta [10]3 years ago

5 0

Mary = M Chris = C
M = 3C-4
M-3=C+9 --> M=C+12
3C-4=C+12
2C=16
C=8
M=3C-4 --> C=8
M=3(8)-4
M=24-4
M=20

Mary is 20, and Chris is 8

You might be interested in

Five less than area of a square is 59 square feet, find the side length of the square

Ne4ueva [31]

Answer:

8 ft

Step-by-step explanation:

Add back those 5 square feet and you find the area of the square to be 64 square feet. Its side length is the square root of that: 8 ft.

8 0

4 years ago

7. (0) + (-9) * (+8) =

ivolga24 [154]

Answer:

-72

Step-by-step explanation:

7 0

3 years ago

I need to find the area of the shaded part. I have the height of the triangle. But I dont know how to get the base of the triang

GalinKa [24]

Well, it is bass times height divided by two

6 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 $x\in A\cap B$ .