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Nesterboy [21]
4 years ago
10

Help with number 23 plz. Plz...

Mathematics
1 answer:
solong [7]4 years ago
7 0

Answer:

D

Step-by-step explanation:

D, it has (x). so its easy to plug in x e.g. 3(4y+6)-2y=3.

Hope this helps plz hit the crown :D

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What is the product of –24 and 35?
beks73 [17]

Answer:

-840

Step-by-step explanation:

"Product" means to multiply, so you multiply -24 by 35. Since -24 has a negative, negative times a positive will equal a negative number.

(Negative times negative is positive.)

(Negative times positive is negative.)

(Positive times positive is obviously positive.)

Using these simple rules, we can do this:

Simply multiply -24 by 35, or you can multiply 24 by 35 which is 840, but DON'T forget to put a negative symbol!

Hope this helps! Let me know if there is anything else.

7 0
3 years ago
Read 2 more answers
Answer this question and show the workout
saul85 [17]

Step-by-step explanation:

Ace:

15800 x (-2/100) = -316

15800 - 316 = 15485

Beta:

17425 x (-3/100) = -522.75

17425 - 522.75 = 16902.25

Carro:

21280 x (-1.5/100) = -319.2

21280 - 319.2 = 20960.8

Delta:

24172 x (-1.8/100) = -435.096

24172 - 435.096 = 23736.90

4 0
4 years ago
A certain freely falling object, released from rest, requires 1.50 s to travel the last 30.0 m before it hit the ground. a.Find
Lyrx [107]
A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2 
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
4 0
4 years ago
How many days are in the year?
seropon [69]
Hey there!


There are 365 days in one year.


Hope I was able to help!
5 0
4 years ago
Read 2 more answers
In the midpoint rule for triple integrals we use a triple riemann sum to approximate a triple integral over a box b, where f(x,
Lana71 [14]
<span>The sub-boxes will have dimensions \frac{2-0}{2} \times \frac{2-0}{2} \times \frac{2-0}{2} =1\times1\times1=1 \ cubic \ units

x sub-intervals are 0 to 1 and 1 to 2. Midpoints are at x= \frac{1}{2} and </span><span>x= \frac{3}{4}
y sub-intervals are 0 to 1 and 1 to 2. Midpoints are at </span><span>y= \frac{1}{2} and </span><span>y= \frac{3}{4}
z sub-intervals are 0 to 1 and 1 to 2. Midpoints are at </span><span>z= \frac{1}{2} and </span><span><span>z= \frac{3}{4}</span>

Let f(x,y,z)=\cos{(xyz)}

\int\limits  \int\limits  \int\limits {f(x,y,z)} \, dV \approx f\left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)+f\left( \frac{1}{2} , \frac{1}{2} , \frac{3}{4} \right)+f\left( \frac{1}{2} , \frac{3}{4} , \frac{1}{2} \right)+f\left( \frac{1}{2} , \frac{3}{4} , \frac{3}{4} \right)
+f\left( \frac{3}{4} , \frac{1}{2} , \frac{1}{2} \right)+f\left( \frac{3}{4} , \frac{1}{2} , \frac{3}{4} \right)+f\left( \frac{3}{4} , \frac{3}{4} , \frac{1}{2} \right)+f\left( \frac{3}{4} , \frac{3}{4} , \frac{3}{4} \right) \\  \\ \approx\cos{ \frac{1}{8} }+\cos{ \frac{3}{16} }+\cos{ \frac{3}{16} }+\cos{ \frac{9}{32} }+\cos{ \frac{3}{16} }+\cos{ \frac{9}{32} }+\cos{ \frac{9}{32} }+\cos{ \frac{27}{64} } \\  \\ \approx0.9922+0.9825+0.9825+0.9607+0.9825+0.9607+0.9607 \\ +0.9123 \\  \\ \approx\bold{7.734}</span>
5 0
3 years ago
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