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n200080 [17]
3 years ago
11

Write in standard form: 4 (2a) + 7 (-4b) + (3.c - 5) * The . is a multiplication sign.

Mathematics
1 answer:
alisha [4.7K]3 years ago
3 0
Refer to cy math.com  they have the answer 
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Louisa needs 3 liters of lemonade and punch for a picnic she has 1800 milliliters of lemonade. How much punch does she need expl
8090 [49]
There are 1000 millimetres in a litre, so 1800 ml is 1.8 litres. 3-1.8 is 1.2.
5 0
3 years ago
Read 2 more answers
Is 1/2 closest to 0 or 1
Irina18 [472]

Answer:

1

Step-by-step explanation:

The reason why is if it’s in the middle you always go for the highest.

7 0
3 years ago
A metal hollow bar whose cross section and dimension are shown below weighs 8x10^3 kg/m^3 and measure 2m in length ..determine t
devlian [24]
1) We calculate the volume of a metal bar (without the hole).

volume=area of hexagon x length
area of hexagon=(3√3 Side²)/2=(3√3(60 cm)²) / 2=9353.07 cm²
9353.07 cm²=9353.07 cm²(1 m² / 10000 cm²)=0.935 m²

Volume=(0.935 m²)(2 m)=1.871 m³

2) we calculate the volume of the parallelepiped

Volume of a parallelepiped= area of the section  x length
area of the section=side²=(40 cm)²=1600 cm²
1600 cm²=(1600 cm²)(1 m² / 10000 cm²=0.16 m²
Volume of a parallelepiped=(0.16 m²)(2 m)=0.32 m³

3) we calculate the volume of a metal hollow bar:
volume of a metal hollow bar=volume of a metal bar -  volume of a parallelepiped

Volume of a metal hollow bar=1.871 m³ - 0.32 m³=1.551 m³

4) we calculate the mass of the metal bar

density=mass/ volume  ⇒ mass=density *volume

Data:
density=8.10³ kg/m³
volume=1.551 m³

mass=(8x10³ Kg/m³ )12. * (1.551 m³)=12.408x10³ Kg

answer: The mas of the metal bar is 12.408x10³ kg  or   12408 kg  


4 0
3 years ago
Simplify sin^2y/sec^2 y−1 to a single trigonometric function
liubo4ka [24]

Answer:

\frac{ { \sin}^{2} y}{ { \sec}^{2}y - 1 }  =  { \cos }^{2} y

Step-by-step explanation:

We know that { \tan }^{2} y =  { \sec }^{2} y - 1

Also , { \tan}^{2} y =   \frac{ { \sin }^{2} y}{ { \cos }^{2}y }

So ,

\frac{ { \sin }^{2}y }{ { \sec }^{2}y - 1 }  =  \frac{ { \sin}^{2} y}{ { \tan }^{2} y}  =  \frac{ { \sin }^{2}y }{ \frac{ { \sin}^{2} y}{ { \cos}^{2}y } }  =  { \cos }^{2} y

6 0
2 years ago
What is the value of StartFraction 1.6 times 10 Superscript 14 Baseline Over 3.2 times 10 Superscript 7 Baseline EndFraction in
Lubov Fominskaja [6]

Answer:

C. 5.0*10 superscript 6

Step-by-step explanation:

i just took the test and i got it right.

8 0
3 years ago
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