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kkurt [141]
3 years ago
5

I need help with this.​

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
4 0

Answer:

DBE = 33

Step-by-step explanation:

add 23 + 27 = 50

subtract 50 from 83

equals 33

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Can someone help me with number 8 please? Thank you.
Naddik [55]
Answer: -4 \le x \le 5
Note: to write the domain in interval notation, you'd write [-4,5]
if you need the domain in set-builder notation, then you'd write \left\{x|x\in\mathbb{R}, \ -4\le x\le 5\right\}

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

Explanation:

The domain is the set of possible x input values. Look at the left most point (-4,-1). The x coordinate here is x = -4. This is the smallest x value allowed. The largest x value allowed is x = 5 for similar reasons, but on the other side of the graph.

So that's how I got -4 \le x \le 5 (x is between -4 and 5; inclusive of both endpoints)

Writing [-4,5] for interval notation tells us that we have an interval from -4 to 5 and we include both endpoints. The square brackets mean "include endpoint"

Writing \left\{x|x\in\mathbb{R}, \ -4\le x\le 5\right\} is the set-builder notation way of expressing the domain. The x\in\mathbb{R} portion means "x is a real number"

6 0
4 years ago
What is the image of (6,-4) after a reflection over the y-axis?
kirill115 [55]

Answer:

(-6,-4)

Step-by-step explanation:

Hope this helps! (:

3 0
3 years ago
If 5=q what is the value of 4q + 2q squared
pentagon [3]

The value of 4q+2q is 30
5 0
3 years ago
Name the sets of numbers to which -7/9 belongs
Arlecino [84]
Ok
the sets are
natural numbers (or counting numbers), this is like 1,2,3,4,5 etc

whole numbers, this is including 0, so 0,1,2,3,4,5,6 etc

integers, this includes the previoius set and negatives, so -3,-2,-1,0,1,2,3,4,5 etc

rational numbers, this is the set of numbers that can be written in form a/b where b≠0, so all integers can be written like this, like example -3=-3/1, so -7/9 belongs here



-7/9 goes in the rational set

5 0
3 years ago
Please HELP ME ON THIS MATH WRITTEN RESPONSE QUESTION.. THANK YOU
sergij07 [2.7K]

Answer:

Quadrilateral ABCD is not a square. The product of slopes of its diagonals is not -1.

Step-by-step explanation:

Point A is (-4,6)

Point B is (-12,-12)

Point C is (6,-18)

Point D is (13,-1)

Given that the diagonals of a square are perpendicular to each other;

We know that the product of slopes of two perpendicular lines is -1.

So, slope(m) of AC × slope(m) of BD should be equal to -1.

Slope of AC = (Change in y-axis) ÷ (Change in x-axis) = (-18 - 6) ÷ (6 - -4) = -24/10 = -2.4

Slope of BD = (Change in y-axis) ÷ (Change in x-axis) = (-1 - -12) ÷ (13 - -12) = 11/25 = 0.44

The product of slope of AC and slope of BD = -2.4 × 0.44 = -1.056

Since the product of slope of AC and slope of BD is not -1 hence AC is not perpendicular to BD thus quadrilateral ABCD is not a square.

4 0
4 years ago
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