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sweet-ann [11.9K]
3 years ago
14

If a quadrilaterals diagonals bisect its vertices, what kind of quadrilateral must it be?

Mathematics
1 answer:
Blababa [14]3 years ago
6 0
<span>Option A, This type of quadrilaterals is a polygon and is called Rhombus because all sides are equal, each pair of acute and obtuse angles are opposite, their diagonals are distinct and perpendicular to each other, bisectors and have an inscribed circumference.</span>
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The sum of five consecutive numbers is -75. Which number is the least of these five numbers?
Ivan
Hello! I can help you with this! Do -75/5 to get -15. -15 will be your middle number. So for consecutive integers, -13 + -14 + -15 + -16 + -17 = 75. Those are five consecutive integers that have a sum of 75. The lowest of 5 integers is -17, because it’s the smallest of them all and the furthest to the left of the number line. A line further to the left of the number line means it’s smaller. Therefore, the rest of the five integers is -17.
6 0
3 years ago
Read 2 more answers
MATH HELP ASAP!! MARKING BRAINLIEST
olga nikolaevna [1]

Answer:

-1, -7, -13, -19

Step-by-step explanation:

-2(-3)-7 =6-7= -1

-2(0)-7 =0-7= -7

-2(3)-7 =-6-7= -13

-2(6)-7 =-12-7= -19

7 0
3 years ago
Read 2 more answers
What is the true solution to 3 In 2+In 8 = 2 In(4x)?<br> 60f<br> Š O OOO<br> TNT 00
alina1380 [7]

Answer:

  x = 2

Step-by-step explanation:

Taking antilogs, you have ...

  2³ × 8 = (4x)²

  64 = 16x²

  x = √(64/16) = √4

  x = 2 . . . . . . . . (the negative square root is not a solution)

___

You can also work more directly with the logs, if you like.

  3·ln(2) +ln(2³) = 2ln(2²x) . . . . . . . . . . . write 4 and 8 as powers of 2

  3·ln(2) +3·ln(2) = 2(2·ln(2) +ln(x)) . . . . use rules of logs to move exponents

  6·ln(2) = 4·ln(2) +2·ln(x) . . . . . . . . . . . . simplify

  2·ln(2) = 2·ln(x) . . . . . . . . . . . subtract 4ln(2)

  ln(2) = ln(x) . . . . . . . . . . . . . . divide by 2

  2 = x . . . . . . . . . . . . . . . . . . . take the antilogs

4 0
3 years ago
Zip codes for the Houston area are five digit numbers with 7 as the first 2 digits of each zip code. The third digit can be 0, 3
djyliett [7]

Answer:

There are 400 possible zip codes in the Houston area

Step-by-step explanation:

Here, we want to calculate the possible number of zip codes in the Houston area

We have 5 digits to form

77 is the first two digits ( this is fixed)

For the third digit, we are selecting 1 number out of 0,3,4 or 5

This means 4 C 1

The remaining digits can be any digits

We have 0-9, a total of 10 digits

The first will be 10 C 1 and the second last digit too is 10 C 1

So the number of possible zip codes will be;

4 C 1 * 10 C 1 * 10 C 1

= 4 * 10 * 10 = 400 possible zip codes

6 0
3 years ago
anna owes home antiques Whose value is 160000 today assuming normal growth. however anna believes the value will grow at 16% per
Ray Of Light [21]
The formula is
A=p (1+r)^t
A future value?
P present value 160000
R interest rate 0.16
T time 3years
A=160,000×(1+0.16)^(3)
A=249,743.36

Use that future value to find the present value at a rate 8% compounded annually
To find p (present value) solve the formula for p
P=A÷ (1+r)^t
Where r is 0.08
P=249,743.36÷(1+0.08)^(3)
p=198,254.33
3 0
3 years ago
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