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3 years ago

What is 10 times x + 17

Mathematics

2 answers:

timofeeve [1]3 years ago

7 0

170! AT LEAST WITH THE PLUS SIGN! :))

Dmitrij [34]3 years ago

4 0

The answer is 170 I think with the plus sign

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Karen has $1.70 in coins. Karen has 8 coins, all of which are quarters or dimes.

Vilka [71]

Answer:

Karen has 6 quarters and 2 dimes

Step-by-step explanation:

Let

x ----> the number of quarter coins Karen has

y ----> the number of dimes coins Karen has

Remember that

$1\ quarter=\$0.25\\1\ dime=\$0.10$

we know that

Equation that represent the amount of coins Karen has

$x+y=8$

isolate the variable y

$y=8-x$ ----> equation A

Equation that represent the value of coins Karen has

$0.25x+0.10y=1.70$ ----> equation B

Solve the system of equations by substitution

substitute equation A in equation B

$0.25x+0.10(8-x)=1.70$

solve for x

$0.25x+0.80-0.10x=1.70$

$0.25x-0.10x=1.70-0.80$

$0.15x=0.90$

$x=6$

Find the value of y

$y=8-x$ ----> $y=8-6=2$

therefore

Karen has 6 quarters and 2 dimes

5 0

3 years ago

22% of _ peaches is 11 peaches

Neko [114]

I’m pretty sure it’s 50 peaches

3 0

3 years ago

Find e^cos(2+3i) as a complex number expressed in Cartesian form.

ozzi

Answer:

The complex number $e^{\cos(2+31)} = \exp(\cos(2+3i))$ has Cartesian form

$\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)$ .

Step-by-step explanation:

First, we need to recall the definition of $\cos z$ when $z$ is a complex number:

$\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}$ .

Then,

$\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}$ . (I)

Now, recall the definition of the complex exponential:

$e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)$ .

So,

$e^{2i-3} = e^{-3}(\cos 2+i\sin 2)$

$e^{-2i+3} = e^{3}(\cos 2-i\sin 2)$ (we use that $\sin(-y)=-\sin y)$ .

Thus,

$e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)$

Now we group conveniently in the above expression:

$e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2$ .

Now, substituting this equality in (I) we get

$\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2$ .

Thus,

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)$

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right]$ .

5 0

3 years ago

What is the value of x?

morpeh [17]

Answer:

x = 130

Step-by-step explanation:

The sum of the exterior angles of a polygon is always 360

140 + x + 56+ 34 = 360

Combine like terms

230 +x = 360

Subtract 230 from each side

230+x-230 = 360-230

x =130

3 0

3 years ago

Read 2 more answers

Helppppppppppppppppppppppppppppppppp

Brrunno [24]

f(x)=x²-1
g(x)=2x-3
f(x)*g(x)=(x²-1)*(2x-3)-----------> this function exists for all real numbers

the answer is the interval (-∞,∞)

using a graph tool see the attached figure

6 0

4 years ago