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3 years ago

HELP ME PLEASEEEEEEEEEEEE

Mathematics

1 answer:

murzikaleks [220]3 years ago

5 0

Answer:

Step-by-step explanation:

if you count the numbers, the longer side is 7 and the other is 5

7+7 and then add 5+5

14+10

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Mr Ordonez gives his math students 34 quizzes during 17 weeks of school. If he gave the same number of quizzes each week, how ma

meriva

2.

Using division we get
34/17= 2

8 0

3 years ago

Read 2 more answers

Given side lengths 4 units, 8 units, and x units, determine the range in which x must lie in order for a triangle to exist.

kvasek [131]

Answer:

C) 4 < x < 12

5 0

3 years ago

Simplify the expression: Show your work. Write the answer in standard form (15 +6d - d²) + (-5d² - 18 + 6d) PLEASE HELP

mezya [45]

Answer:

-6d^2 + 12d - 3

Step-by-step explanation:

use PEMDAS

parentheses

exponents

multiplication

division

addition

subtraction

use that order to solve problems like that.

6 0

3 years ago

How do you solve 1-5+1×(4×4-31)×8

Alja [10]

Hey there!

In order to solve this, remember PEMDAS. First, start with solving whatever is in parentheses. Then, move on to exponents. Then, complete the multiplication and division. Lastly, add or subtract anything remaining. Omit any steps that aren't present.

Parentheses:

<span>1 – 5 + 1 × (4 × 4 – 31) × 8
</span>1 – 5 + 1 × (–15) × 8

Multiplication/Division (Left to Right):

1 – 5 + (–15) × 8
1 – 5 + (–120)
1 – 5 – 120

Addition/Subtraction (Left to Right):

–4 – 120
–124

Your answer is –124.

Hope this helped you out! :-)

5 0

3 years ago

Determine the volume of the parallelepiped with one vertex at the origin and the three vertices adjacent to it at (2, â1, â1), (

valentinak56 [21]

Answer:

Step-by-step explanation:

Here is the complete question

Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at (1, 0, -3), (1, 2, 4), and (5, 1, 0).

Solution

We find the volume of the parallelepiped by making a 3 × 3 column matrix whose columns are the corresponding coordinates of the vertices of the parallelepiped.

So, (1, 0, -3), (1, 2, 4) and (5, 1, 0)

$A = \left[\begin{array}{ccc}1&1&5\\0&2&1\\-3&4&0\end{array}\right]$

The determinant of A is the volume of the parallelepiped. So,

detA = 1(2 × 0 - 4 × 1) - 1(0 × 0 - (-3) × 1) + 5(0 × 4 - (-3) × 2)

= 1(0 - 4) - 1(0 + 3) + 5(0 + 6)

= 1(-4) - 1(3) + 5(6)

= -4 - 3 + 30

= 23

So the volume of the parallelepiped is 23

4 0

2 years ago