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3 years ago

11

Brynne has 13 books she has 8 more books than games how many game does Brynne have

2 answers:

GalinKa [24]3 years ago

7 0

The answer is she has 5 more books

vagabundo [1.1K]3 years ago

3 0

Answer:

5

Step-by-step explanation:

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I have just one question, could someone please answer it?? (see linked file)

Annette [7]

The solution is x=1 and y=-2

Further explanation:

Given equations are:

$\frac{7}{2}x-\frac{1}{2}y=\frac{9}{2}\\3x-y=5$

Multiplying equation no. 1 with 2 will give us:

$7x-y=9$ Eqn 3

From equation no 2:

$3x-y=5\\-y=5-3x\\y=3x-5$

Putting the value of y in eqn 3

$7x-(3x-5)=9\\7x-3x+5=9\\4x=9-5\\4x=4\\\frac{4x}{4}=\frac{4}{4}\\x=1\\Putting\ x=1\ in\ equation\ 2\\3(1)-y=5\\3-y=5\\-y=5-3\\-y=2\\y=-2\\$

The solution is x=1 and y=-2

Keywords: Linear equations, Solution set

Learn more about linear equations at:

#LearnwithBrainly

8 0

3 years ago

2 3/7 + (-1 3/4) i need help please

julsineya [31]

Answer:

1/28

Step-by-step explanation:

7 0

3 years ago

Would you please help me with this quiz answer, I will really appreciate if you do.

laila [671]

Answer:

a . domain 5,0,7,9,0

range -2,-2,-4,8,2

b. domain 2,4,8,9

range 1,2,4,11

Step-by-step explanation:

<h3>a is not a function</h3>

because function is a relationship in which each domain element occurs only once.

<h3>b is a function</h3>

6 0

2 years ago

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

7 0

3 years ago

Ill give u brainiest for correct ans

Marrrta [24]

Answer:

Step-by-step explanation:

7 0

3 years ago