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2 years ago

Write four different equations that have –3 as the solution. (2 points each)

Mathematics

2 answers:

marissa [1.9K]2 years ago

8 0

Answer:

equation 1: x-5=-8 <=> x=-3

equation 2: x-3=-6 <=> x=-3

equation 3: x.3=-9 <=> x=-3

equation 4: -9 : x=3 <=> x=-3

=> Four equations that have –3 as the solution: x=-3

Likurg_2 [28]2 years ago

6 0

-2-1=-3 -6+3=-3 -12+9=-3

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Solve the equation for all values of x by completing the square. x2 + 8.0 + 15 = 0

Aleonysh [2.5K]

Answer:

x= -3

x= -5

Step-by-step explanation:

x^2 + 8x + 15 = 0

find which 2 numbers when multiplied equal 15 and when added equal 8

they are 3 and 5

(x+3)(x+5)

x+3 =0

x+5=0

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6 0

2 years ago

An illusionist needs up to 10 volunteers for a show. She needs no fewer than 4 female volunteers. Let x represent the number of

Sliva [168]

The correct answers are:

B- x + y ≤ 10 ; and D- x ≥ 4

Explanation:

The illusionist needs at least 4 female volunteers. This means he could use 4 or more; this means x, the number of female volunteers, is greater than or equal to 4:

x ≥ 4

The total of number of volunteers needed is up to 10. This means the illusionist can use 10 or fewer people; this means the total is less than or equal to 10.

The number of female volunteers is x and the number of male volunteers is y; together they create the expression x+y. This gives us

x+y ≤ 10

7 0

3 years ago

Read 2 more answers

2^n=8^20. Solve for n

damaskus [11]

Hi the answer is 60 but the picture will show you the work!!

3 0

3 years ago

Read 2 more answers

1/3÷100 will the answer be smaller than 1 close to 1 or much greater than 1

Radda [10]

Answer:

It will be smaller than one

Step-by-step explanation:

because i said so

7 0

2 years ago

The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00. Assu

m_a_m_a [10]

Answer: 0.0170

Step-by-step explanation:

Given : The mean amount purchased by a typical customer at Churchill’s Grocery Store is $23.50, with a standard deviation of $5.00.

i.e. $\mu=23.50$

$\sigma=5$

We assume the distribution of amounts purchased follows the normal distribution.

Sample size : n=50

Let $\overline{x}$ be the sample mean.

Formula : $z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

Then, the probability that the sample mean is at least $25.00 will be :-

$P(\overline{x}\geq\25.00)=P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\geq\dfrac{25-23.50}{\dfrac{5}{\sqrt{50}}})\\\\=P(z\geq2.12)\\\\=1-P(z$

Hence, the likelihood the sample mean is at least $25.00= 0.0170

5 0

3 years ago