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7 months ago

Please answer need to turn it in please help me answer

Mathematics

1 answer:

Elena-2011 [213]7 months ago

4 0

Answer: Sir in what way would you like me to explain this

Step-by-step explanation:

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I'm bad word problems can you help me

IRINA_888 [86]

Use 12 into the equation 2+0.35m

2 + 0.35(12)

Follow PEMDAS

P= Parentheses

E= Exponents

M= Multiplication

A= Addition

S= Subtraction

2+0.35(12)

= 2 +4.2

= 6.20

Answer: $6.20

3 0

3 years ago

One serving of a certain brand of yogurt contains 12 G of fat the low-fat version contains 38.75% less fat per serving how many

Mkey [24]

Answer:

One serving of law-fat yogurt contains 7.35 g of fat.

Step-by-step explanation:

One serving of

a certain brand of yogurt contains 12 g of fat
the low-fat yogurt contains 38.75% less fat

Let x g be the amount of fat in low-fat yogurt. Then

12 g of fat - 100%

x g of fat - 100% - 38.75% = 61.26%

Write a proportion

$\dfrac{12}{x}=\dfrac{100}{61.25}$

Cross multuply

$100\cdot x=12\cdot 61.25\\ \\100x=735\\ \\x=\dfrac{735}{100}=7.35$

One serving of law-fat yogurt contains 7.35 g of fat.

3 0

3 years ago

Anyone know this geometry question?

mr Goodwill [35]

Given:
∠JEK + ∠DFH = 90°

Find:
relationship, if any, of ∠JEK to angles CFL, HFG, LFE

Solution:
We note that ∠DFG = 90° = ∠DFH + ∠HFG, so ∠JEK ≅∠HFG.
We also note that ∠HFG and ∠CFL are vertical angles, so ∠JEK ≅∠CFL.
And ∠LFE and ∠DFH are vertical angles, so ∠JEK + ∠LFE = 90°.

Evaluating the choices, we find
A) false. The angles are not complementary, they are equal.
B) false. The angles are not equal, they are complementary.
C) false. The angles are not supplementary, they are equal.
D) TRUE. The angles are equal.

The appropriate choice is ...
D) ∠JEK ≅∠CFL

6 0

2 years ago

There are 3 islands A,B,C. Island B is east of island A, 8 miles away. Island C is northeast of A, 5 miles away and northwest of

Nostrana [21]

Answer:

The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

$AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta$ (1)

Where:

$AC$ - The distance from A to C, measured in miles.

$AB$ - The distance from A to B, measured in miles.

$BC$ - The distance from B to C, measured in miles.

$\theta$ - Bearing from island B to island C, measured in sexagesimal degrees.

Then, we clear the bearing angle within the equation:

$AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta$

$\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}$

$\theta = \cos^{-1}\left(\frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC} \right)$ (2)

If we know that $BC = 7\,mi$ , $AB = 8\,mi$ , $AC = 5\,mi$ , then the bearing from island B to island C:

$\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]$

$\theta \approx 38.213^{\circ}$

The bearing needed to navigate from island B to island C is approximately 38.213º.

8 0

2 years ago

Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p

Juli2301 [7.4K]

I'm reading this as

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy$

with $\nabla f=(2xe^{-y},2y-x^2e^{-y})$ .

The value of the integral will be independent of the path if we can find a function $f(x,y)$ that satisfies the gradient equation above.

You have

$\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}$

Integrate $\dfrac{\partial f}{\partial x}$ with respect to $x$ . You get

$\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx$
$f=x^2e^{-y}+g(y)$

Differentiate with respect to $y$ . You get

$\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]$
$2y-x^2e^{-y}=-x^2e^{-y}+g'(y)$
$2y=g'(y)$

Integrate both sides with respect to $y$ to arrive at

$\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy$
$y^2=g(y)+C$
$g(y)=y^2+C$

So you have

$f(x,y)=x^2e^{-y}+y^2+C$

The gradient is continuous for all $x,y$ , so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e$

8 0

2 years ago