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3 years ago

Which pointis not part of the solution of the inequality y ≥ -|x-4|-3

Mathematics

1 answer:

amid [387]3 years ago

7 0

Answer:

(4,-4)

Step-by-step explanation:

To find the point which is not a solution, identify the point which is not in the set of solutions. This will be a point which is not in the shaded section of its graph. Graph the inequality by shifting the parent function down 3 units and to the left 4 units. Then flip the graph because of the negative out front. See the attached graph. All the points lie in the shaded except for (4,-4).

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Evaluate another function please

kari74 [83]

Answer:

70, I think

Step-by-step explanation:

f(10)= -10^2- 3(10)

f(10)= 100-30

f(10) = 70

6 0

2 years ago

2/3 + (-5/7) .........

dmitriy555 [2]

Answer:

-1/21

Step-by-step explanation:

2/3 + -5/7

Get a common denominator of 21

2/3 *7/7 = 14/21

-5/7*3/3 = -15/21

14/21 - 15/21 = -1/21

4 0

3 years ago

Read 2 more answers

Solve the system of equations using matrices. Use the Gauss- Jordan elimination method And find a solution set

KATRIN_1 [288]

$\begin{gathered} x+y+z=4 \\ x-y-z=0 \\ x-y+z=8 \\ \text{The system using matrix is} \\ \begin{bmatrix}{1} & {1} & {1}, & {4} \\ {1} & {-1} & {-1,} & {0} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix}\rightarrow F2=F2-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix} \\ \rightarrow F3=F3-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F2=-\frac{1}{2}F2 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F3=F3+2F2=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {2,} & {8}{}{}\end{bmatrix}\rightarrow F3=\frac{1}{2}F3 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F2=F2-F3=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \rightarrow F1=F1-F3=\begin{bmatrix}{1} & {1} & {0}, & {0} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F1=F1-F2 \\ =\begin{bmatrix}{1} & {0} & {0}, & {2} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \text{Therefore, the solution is }x=2,\text{ y=-2 and z=4} \end{gathered}$

6 0

1 year ago

The table gives the relative frequencies of recipes that contains sugar and salt, contain at least one of those ingredients, or

tigry1 [53]

Answer-

The probability that a randomly selected recipe does not contain sugar, given that it contains salt is 22.4%

Solution-

The given table in the link shows the relative frequencies of recipes that contains sugar and salt, or contains at least one of those ingredients, or contains neither of those ingredients.

We have to find the conditional probability that the recipe doesn't contain sugar, given that it contains salt.

We know that, the conditional probability of occurrence of A given that B occurs is,

$P(A|B)=\frac{P(A\ and\ B)}{P(B)} =\frac{P(A\bigcap B)}{P(B)}$

$P(\text{Doesn't contain sugar}\ |\ \text{Contains salt})=\frac{P(\text{Doesn't contain sugar}\ and\ \text{Contains salt})}{P(\text{Contains salt})}$

$P(\text{Doesn't contain sugar}\ and\ \text{Contains salt})=\frac{0.15}{1} =0.15\\P(\text{Contains salt})=\frac{0.67}{1}=0.67$

Putting these values,

$P(\text{Doesn't contain sugar}\ |\ \text{Contains salt})=\frac{0.15}{0.67} =0.224=22.4\%$

5 0

3 years ago

A company makes gift boxes in different sizes following the pattern shown below. What is the volume of the fourth gift box to th

AVprozaik [17]

The volume of the fourth box is of 410.1 cubic inches.

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The volume of a rectangular prism is given by the base area multiplied by the height, that is:

$V = A_bh$

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For the first three boxes, the height is constant at 9 inches.
At the base, the edge length is multiplied by 1.5 each time, as $\frac{4.5}{3} = \frac{3}{2} = 1.5$ , thus, for the fourth box, the edge length, in inches, will be of $4.5 \times 1.5 = 6.75$

---------------------

Square base, of length 6.75 inches, thus:

$A_b = 6.75^2 = 45.5625$

Height of 9 inches, thus $h = 9$ and:

$V = A_bh = 45.5625(9) = 410.1$

The volume of the fourth gift box will be of 410.1 cubic inches.

A similar problem is given at brainly.com/question/23756783

4 0

2 years ago