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3 years ago

What is the value of the expression? 2(-5.25)

Mathematics

1 answer:

Alexus [3.1K]3 years ago

6 0

Answer:

-10.5

Step-by-step explanation:

2 x (-5.25) = -10.5

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Its me arg i need help

Alex787 [66]

Answer:

x=8

Step-by-step explanation:

so since the whole thing should equal 180, and the right triangle is 90,

(3x+11)+(5x+15)=90 then solve for that

8x+26=90

8x=64

divide both sides by 8

so x=8

then plug in 8 for the xs so c is 3(8)+11 --> 24+11 so m<C=35*

m<D=90*

5(8)+15 40+15 so m<E=55

7 0

2 years ago

Read 2 more answers

Use the percent proportion to solve for X. 16 is 32% of x. Find the value of 1.

Umnica [9.8K]

Answer:

I dont get the question, is there suposed to be a chart?

Step-by-step explanation:

7 0

3 years ago

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

<u>Answer-</u>

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

<u>Solution-</u>

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$

Now, equating this to 0

$(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x}) =0$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}-(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})$

$\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}$

$\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}$

$\Rightarrow 2304e^{2x}-16e^{2x}-1=0$

Solving this eq,

we get $x= \frac{1}{2304e^4-16e^2}$

∴ At $x= \frac{1}{2304e^4-16e^2}$ the curvature is maximum.

6 0

2 years ago

Write the following fraction as tenths in fraction and decimals.4/10

Veseljchak [2.6K]

If you wanted to write 4/10 as a decimal it would be .4

3 0

3 years ago

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Let T:R²->R² be a linear transformation ,and assume that T (1,2)=(-1,1) and T(1,-1)=(2,3)

zavuch27 [327]

Answer:

(-4,-1)

Step-by-step explanation:

We are given T(1,2)=(-1,1) and T(1,-1)=(2,3) and T is a linear transformation.

This implies for scalars a and b that

T(a(1,2)+b(-1,1))=aT(1,2)+bT(-1,1)

T((a,2a)+(-b,b))=a(-1,1)+b(2,3)

T((a-b,2a+b))=(-a,a)+(2b,3b)

T((a-b,2a+b))=(-a+2b,a+3b)

This means we should be able to solve the system below to find a and b for T(3,3):

a-b=3 and 2a+b=3

Add equations to eliminate b and solve for a:

3a=6

Divide 3 on both sides:

a=2

If a-b=3 and a=2, then b=-1.

Plug in a=2, b=-1:

T((a-b,2a+b))=(-a+2b,a+3b)

T((2--1,2×2+-1)=(-2+2×-1,2+3×-1)

T(3,3)=(-4,-1).

4 0

2 years ago