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Viefleur [7K]
3 years ago
14

Please help find which pair works for this equation

Mathematics
1 answer:
JulijaS [17]3 years ago
4 0

Answer:

h = 20 and a = 6

Step-by-step explanation:

Substitute the values of a into the tight side of the inequality and check the validity of the result against the given value of h

a = 1 → 4(1) - 3 = 4 - 3 = 1 and 2 > 1

a = 2 → 4(2) - 3 = 8 - 3 = 5 and 6 > 5

a = 3 → 4(3) - 3 = 12 - 3 = 9 and 10 > 9

a = 6 → 4(6) - 3 = 24 - 3 = 21 and 20 < 21

h = 20 and a = 6 are the only pair that satisfy the inequality

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F(x) = -x + 4 and g(x) = 6x + 3, find (f(x) - g(x))<br><br>A. -5x-1<br>B. 5x +1<br>C. -7x +1<br>​
Pepsi [2]

Answer:

C

Step-by-step explanation:

f(x) - g(x)

= - x + 4 - (6x + 3)

= - x + 4 - 6x - 3 ← collect like terms

= - 7x + 1 → C

8 0
3 years ago
In one baseball season, Peter hit twice the difference of the number of home runs Alice hit and 6. Altogether, they hit 18 home
saw5 [17]
Below is the solution:

There are two equations to be solved here.  The first one we assign variables to the home runs Peter hits, P, and Alice hits, A. P=2*(A-6) The second equation is the sum of both players home runs.P + A = 18Solving for P yieldsP=18-A We substitute the solution for P into the first equation and solve for A.(18-A) = 2*(A-6)18 - A = 2A - 123A = 30A = 10 Now that A is known, we can plug it into either equations for P to find how many home runs Peter hitsP=18 - (10) = 8orP=2((10)-4) = 8
7 0
3 years ago
One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R
marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

Step-by-step explanation:

Given

Urn\ 1 = \{B_1, R_1, R_2, R_3\}

Urn\ 2 = \{R_4, R_5, B_2, B_3\}

Solving (a): A possibility tree

If urn 1 is selected, the following selection exists:

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

If urn 2 is selected, the following selection exists:

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

<em>See attachment for possibility tree</em>

Solving (b): The total number of outcome

<u>For urn 1</u>

There are 4 balls in urn 1

n = \{B_1,R_1,R_2,R_3\}

Each of the balls has 3 subsets. i.e.

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

So, the selection is:

Urn\ 1 = 4 * 3

Urn\ 1 = 12

<u>For urn 2</u>

There are 4 balls in urn 2

n = \{B_2,B_3,R_4,R_5\}

Each of the balls has 3 subsets. i.e.

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

So, the selection is:

Urn\ 2 = 4 * 3

Urn\ 2 = 12

Total number of outcomes is:

Total = Urn\ 1 + Urn\ 2

Total = 12 + 12

Total = 24

5 0
3 years ago
Using appropriate properties find: 2/3 × 3/4 + 3/7 × 3/4​
Viktor [21]

Answer:

\frac{23}{28}

Step-by-step explanation:

\frac{2}{3}  \times  \frac{3}{4}  +  \frac{3}{7}  \times  \frac{3}{4}

=  >  \frac{3}{4} ( \frac{2}{3}  +  \frac{3}{7} )

=  >  \frac{3}{4}  \times  \frac{23}{21}

Reducing 3 from numerator and denominator,

=  >  \frac{23}{4 \times 7}  =  \frac{23}{28}

3 0
3 years ago
Choose the value that breaks this gem into 10 equal pieces
AnnyKZ [126]

Answer:

25 percent

Step-by-step explanation:

cuz am smart

7 0
4 years ago
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