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3 years ago

Please show all work cuase I don’t understand thank you

Mathematics

1 answer:

Natali [406]3 years ago

5 0

Answer:

13a. 0.92 miles

13b. C

Step-by-step explanation:

13a. The first ride costs $2.75 more than the second ride. Each quarter mile after the first quarter mile costs $0.75, so the difference in distance is:

0.75 d = 2.75

d = 3.67

The difference is 3.67 quarter miles, or 0.92 miles.

13b. The total cost is the cost of the first quarter mile plus the cost of the rest of the quarter miles.

C(n) = 2.50 + 0.75 (n − 1)

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X^2-x+2=0 Please use the quadratic formula

Ainat [17]

Answer:

$\boxed{\sf x=\cfrac{1\pm \sqrt{-7} }{2}}$

Step-by-step explanation:

$\sf x^2-x+2=0$

Use quadratic formula:

$\boxed{\sf \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

$\sf a=1, b=-1, c=2$

___________________________

$\sf x_1,_2=\cfrac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\times \:1\times \:2}}{2\times \:1}$

$\sf \sqrt{\left(-1\right)^2-4\times \:1\times \:2}$

$\sf =\sqrt{-7}$

_________________

$\sf x_1,_2\cfrac{-\left(-1\right)\pm \sqrt{-7}}{2\times \:1}$

$\sf -(-1)=1$

$2\times 1=2$

$\sf x_1,_2=\cfrac{1\pm \sqrt{-7} }{2}$

_______________________________________

Learn more: brainly.com/question/22286698

4 0

2 years ago

Please please help me

enyata [817]

The area of a parallelogram is given by

$A=bh$

Now, if we consider the 9.9 inches side as the base, then the height is the one labeled with 5.5 inches.

If instead we choose the 11 inches side as the base, the height is h.

So, we can express the area in this two equivalent ways:

$A = 9.9\cdot 5.5 = 11\cdot h$

Solving for h, we have

$h = \dfrac{9.9\cdot 5.5}{11} = 4.95$

8 0

3 years ago

X/4 - y/3 =1 X and Y axis standard form

Semenov [28]

Answer:

Step-by-step explanation:

x/4 - y/3 = 1..multiply everything by the LCD....which is 12

3x - 4y = 12 <== standard form

=====================================

if it makes it easier for u, x/4 is the same as (1/4)x and y/3 is the same as (1/3)y

so ur problem can be written as : (1/4)x - (1/3)y = 1....multiply by 12

12(1/4)x - 12(1/3)y = 12(1)

(12/4)x - (12/3)y = 12.....reduce

3x - 4y = 12

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3 years ago

Round the quantity to the given place value. 382.993 to nearest hundredth

adoni [48]

Answer: It is (383,000)

Hope this helps, have a great day/night!

~IFoundJiminsLostJams~

Step-by-step explanation:

3 0

2 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

3 years ago