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Hatshy [7]
3 years ago
7

Which of the following could be units for mass? (Select ALL that are correct) *

Mathematics
1 answer:
djyliett [7]3 years ago
5 0
I only know that kg is for mass
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The width of a rectangle is 11 inches more than 4 times the length. if x represents the length, write an algebraic expression in
Aleonysh [2.5K]
It's always easier to start by drawing a diagram.
We can label one side X. Because it's a rectangle, the side opposite side X must also be X.

Now we need to label the other two sides. These sides are 11 more than 4 times the length X. 4 times X is expressed as 4X. Now we just add 11 to that, and we get 4X + 11.
Now there are two ways to go about writing an algebraic expression (ie equation) for the perimeter.

The first way is by using brackets.
Because there are 2 sides X, we can already jot down these two sides simply as 2X.
Now here's the fun bit. We can chuck some brackets around the width and then write 2 in front. This is 2(4X + 11). This essentially means 2 lots of the width, which is what we are after.
Now we can write that the perimeter, P, is equal to 2X + 2(4X +11).
Now we can simplify this expression by expanding it. Let's temporarily forget the 2X for a moment and focus on the 2(4X + 11).
To expand it out, we have to multiply everything within the brackets by the number on the outside. So we have to go 4X × 2, and we have to go 2 × 11. Remembering like terms, we end up with 8X + 22.
Now we can rephrase the equation as
P = 2X + 8X + 22
Using like terms, we can simplify this one step further by adding 2X and 8X together to give us 10X.
And now, the final equation:
P = 10X + 22

The other way requires no brackets, just a bit of counting. You may find this easier.
You literally just write out each side and set it equal to the perimeter, P.
Then you identify the like terms and add them together. I have coloured them for clarity.
Then you will get your final expression.

If you are thinking about going onto doing harder maths in the future, I suggest you learn how to use the brackets method, as it will come in handy for those tricky questions. However, if you're just looking to scrape a pass, then the easier method should do you just fine :)

Hope I helped

3 0
4 years ago
Read 2 more answers
The base of a triangular prism is an isosceles right triangle with a hypotenuse of 32−−√ centimeters. The height of the prism is
Roman55 [17]
Isosceles triangle: two equal sides.
 We have the following relationship:
 root (32) = root (L ^ 2 + L ^ 2)
 root (32) = root (2L ^ 2)
 root (32) = Lraiz (2)
 root (32) / root (2) = L
 The surface area is:
 Area of the base and top:
 A1 = (1/2) * (root (32) / root (2)) * (root (32) / root (2))
 A1 = (1/2) * (32/2)
 A1 = (1/2) * (16)
 A1 = 8
 Area of the rectangles of equal sides:
 A2 = (root (32) / root (2)) * (6)
 A2 = 24
 Rectangle area of different side:
 A3 = (root (32)) * (6)
 A3 = 33.9411255
 The area is:
 A = 2 * A1 + 2 * A2 + A3
 A = 2 * (8) + 2 * (24) + (33.9411255)
 A = 97.9411255
 Round to the nearest tenth:
 A = 97.9 cm
 Answer:
 The surface area of the triangular prism is:
 
A = 97.9 cm
3 0
4 years ago
Given these values<br> f(x)=x^2 -3x-4, g(x)=-2x, h(x)-x^3<br><br> Evaluate <br> (f(g(h(x)))
goldfiish [28.3K]

\begin{cases} f(x)=x^2-3x-4\\ g(x)=-2x\\ h(x)=-x^3 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{g(~~h(x)~~)}{g(~~-x^3~~)}=-2\boxed{-x^3}\implies g(~~-x^3~~)=2x^3 \\\\[-0.35em] ~\dotfill\\\\ \underset{f(~~g(~~-x^3~~)~~)}{\stackrel{f(~~g(~~h(x)~~)~~)}{f(~~2x^3~~)}}=\boxed{2x^3}^2-3\boxed{2x^3}-4\implies f(~~2x^3~~)=2^2x^{3\cdot 2}-6x^3-4 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{f(~~g(~~h(x)~~)~~)}{f(~~2x^3~~)}=4x^6-6x^3-4~\hfill

4 0
2 years ago
1. FLVS
Tom [10]
X is -8 in the vertical asymptote

3 0
3 years ago
Evaluate ∫SF⃗ ⋅dA⃗ , where F⃗ =(bx/a)i⃗ +(ay/b)j⃗ and S is the elliptic cylinder oriented away from the z-axis, and given by x2/
Norma-Jean [14]

Answer:

Therefore surface integral is \pi(a^2+b^2)c-0-0=\pi(a^2+b^2)c.

Step-by-step explanation:

Given function is,

\vec{F}=\frac{bx}{a}\uvec{i}+\frac{ay}{b}\uvec{j}

To find,

\int\int_{S}\vec{F}dS  

where S=A=surfece of elliptic cylinder we have to apply Divergence theorem so that,

\int\int_{S}\vec{F}dS

=\int\int\int_V\nabla.\vec{F}dV

=\int\int\int_V(\frac{b}{a}+\frac{a}{b})dV  

=\frac{a^2+b^2}{ab}\int\int\int_VdV

=\frac{a^2+b^2}{ab}\times \textit{Volume of the elliptic cylinder}

=\frac{a^2+b^2}{ab}\times \pi ab\times 2c=\pi (a^2+b^2)c

  • If unit vector \cap{n} directed in positive (outward) direction then z=c and,

\int\int_{S_1}\vex{F}.dS_1=\int\int_{S_1} . dA      

=\int\int_{S_1}.dA=0

  • If unit vector \cap{n} directed in negative (inward) direction then z=-c and,

\int\int_{S_2}\vex{F}.dS_2=\int\int_{S_2}. -dA      

=\int\int_{S_2}. -dA=0

Therefore surface integral without unit vector of the surface is,

\pi(a^2+b^2)c-0-0=\pi(a^2+b^2)c

5 0
4 years ago
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