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SCORPION-xisa [38]

4 years ago

13

Layla answer 21 of the 25 questions on his history test correctly.What decimal represents the fraction of problem he answer inco

rrectly.

1 answer:

valkas [14]4 years ago

7 0

0.84 would be the decimal

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Using the geometric mean and Pythagorean theorem, calculate the values of the missing sides. Round your answers to the thousandt

Pachacha [2.7K]

Answer:

a = 9.849

b = 20.25

c = 491.03

Step-by-step explanation:

By using Pythagoras theorem in the right triangle BDC,

(Hypotenuse)² = (Leg 1)² + (Leg 2)²

BC² = BD² + DC²

a² = 9² + 4²

a = $\sqrt{(81+16)}$

a = $\sqrt{97}$

a = 9.8489

a ≈ 9.849 units

By mean proportional theorem,

$\frac{\text{DC}}{\text{BD}}=\frac{\text{BD}}{\text{AD}}$

AD × DC = BD²

b × 4 = 9²

b = $\frac{81}{4}$

b = 20.25 units

BY Pythagoras theorem in ΔADB,

AB² = AD² + BD²

c² = b² + 9²

c² = (20.25)² + 9²

c² = 410.0625 + 81

c = 491.0625

c = 491. 063 units

6 0

3 years ago

Describe the error in the subtraction shown below

Gemiola [76]

Answer:

-11x+12

Step-by-step explanation:

(x^2-4x+3)-(x^2+7x-9)

x^2-x^2-4x-7x+3-(-9)

-11x+3+9

-11x+12

8 0

3 years ago

Read 2 more answers

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

What is the equation of the line that passes through the point (-4,-3) and has a slope of 5

lbvjy [14]

Answer:

5x-y=-17

Step-by-step explanation:

that is the answer

7 0

3 years ago

What are the solutions of x^2+5x=2? Check all that apply

astraxan [27]

X^2+5x=2
you then solve by completing the square using the formula (b/2)^2 in order to create the new term. Solve for x by using this term to complete the square.
Your answer is then
x=-5/2+-√33/2

3 0

3 years ago

Read 2 more answers