I need this done today whoever gets this correct gets brainliest!!! PLZZ HELPP MEE!!! hurryy!!

Mathematics

1 answer:

kenny6666 [7]3 years ago

3 0

Answer:

i dont know if this is right but 0.125 for every one cup i got this by dividing the fractions

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Four decimal numbers that are less than 1.004

shusha [124]

Answer:

1.003

1.002

1.001

1.000

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The first term of an arithmetic sequence is -3 and the fifteenth term is 53. What is the common difference of the sequence?

vampirchik [111]

So, from the first term, to the fifteen term, there are 14 terms to go through.

that means, if say the common difference is say "d", we have to add "d" 14 times, to get to the 15th term, lemme make a quick table,

$\bf \begin{array}{llll}
term&value\\
----&------\\
1&-3\\
2&-3+d\\
3&-3+d+d\\
4&-3+d+d+d\\
5&-3+d+d+d+d\\
6&-3+d+d+d+d+d\\
7&-3+d+d+d+d+d+d\\
8&-3+d+d+d+d+d+d+d\\
9&-3+d+d+d+d+d+d+d+d\\
10&-3+d+d+d+d+d+d+d+d+d\\
11&-3+d+d+d+d+d+d+d+d+d+d\\
12&-3+d+d+d+d+d+d+d+d+d+d+d\\
13&-3+d+d+d+d+d+d+d+d+d+d+d+d\\
14&-3+d+d+d+d+d+d+d+d+d+d+d+d+d\\
15&-3+d+d+d+d+d+d+d+d+d+d+d+d+d+d\\
&53
\end{array}\\\\
-------------------------------\\\\
-3+14d=53\implies 14d=56\implies d=\cfrac{56}{14}$

and surely you know how much that is.

4 0

3 years ago

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!