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Anuta_ua [19.1K]
3 years ago
15

Is (3,10) a solution to this system of equations? y=2x+3 y=x+7

Mathematics
1 answer:
Ksju [112]3 years ago
4 0

Answer: no

Step-by-step explanation:

plug in (3,10) to y=2x+3

10=2(3)+3

10=6+3

10\neq9

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Which of the following functions is graphed below?
denis-greek [22]

The option that is the functions is graphed below is known as option A = Y= (x+4) -2.

<h3>What is the function about?</h3>

The function graphed above is Y=|x+4|-2 because  when you solve for the graphs,  the number in the x-axis is always changed or turned to  the other side and there one can see that it would be -4 would be +4, and +4 and also -4.

Note that

Y = (x+4) -2.

x = (0 + 4) - 2

x = 4-2

x =2

Therefore, plug x into the equation.

Y = (x+4) -2.

= ( 2 + 4) - 2

= (6) -2

= 4

Therefore, The option that is the functions is graphed below is known as option A = Y= (x+4) -2.

See full question in the image attached.

Learn more about functions  from

brainly.com/question/25638609

#SPJ1

5 0
2 years ago
Answer this question please and thank you
Murljashka [212]

Answer:

1

1

0

0

Step-by-step explanation:

here´s and example for 1

1⁵

1x1x1x1x1=1

so 1 raised to the power of any number is 1

1 multiply by itself any number of times is 1

here´s an example for 0

0⁴

0x0x0x0=0

hence, the vale of 0 raised to any power is 0.

0 multiplied by itself any number of times is 0

8 0
3 years ago
Please help help help help help please ASAP
lora16 [44]

Answer

~~~~~~~~~~~~

slope: 1

y intercept: 4

equation: y=x+4

6 0
3 years ago
Is there a answer to this
hodyreva [135]

Answer:

mean=71

Step-by-step explanation:

Mean x¯¯¯x¯

71

Median x˜x~

72

Mode

64, 83, 79, 38, 81, 90, 59, 93, 70, 73, 71, 51

Range

55

Minimum

38

Maximum

93

Count n

12

Sum

852

Quartiles

Quartiles:

Q1 --> 61.5

Q2 --> 72

Q3 --> 82

Interquartile

Range IQR

20.5

Outliers

none

5 0
3 years ago
How many integers between 10000 and 99999, inclusive, are divisible by 3 or<br> 5 or 7?
Yuki888 [10]

Answer: Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.

Step-by-step explanation:

Since we have given that

Integers between 10000 and 99999 = 99999-10000+1=90000

n( divisible by 3) = \dfrac{90000}{3}=30000

n( divisible by 5) = \dfrac{90000}{5}=18000

n( divisible by 7) = \dfrac{90000}{7}=12857.14

n( divisible by 3 and 5) = n(3∩5)=\dfrac{90000}{15}=6000

n( divisible by 5 and 7) = n(5∩7) = \dfrac{90000}{35}=2571.42

n( divisible by 3 and 7) = n(3∩7) = \dfrac{90000}{21}=4285.71

n( divisible by 3,5 and 7) = n(3∩5∩7) = \dfrac{90000}{105}=857.14

As we know the formula,

n(3∪5∪7)=n(3)+n(5)+n(7)-n(3∩5)-n(5∩7)-n(3∩7)+n(3∩5∩7)

=30000+18000+12857.14-6000-2571.42-4258.71+857.14\\\\=48884.15

Hence, there are approximately 48884 integers are divisible by 3 or 5 or 7.

5 0
4 years ago
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