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2 years ago

Fill in the blanks -.-

Mathematics

2 answers:

RideAnS [48]2 years ago

7 0

The anwser is um 5 :)

Verdich [7]2 years ago

3 0

I’m guessing it’s 5 because in ounces it says 1,3 they skipped 2 soooo.

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I GIVEEE BRAINLILSTT

alekssr [168]

Answer:

(x, y) --> (x + 14, y + 8)

Step-by-step explanation:

Look at 1 original point and its corresponding translated point.

Let's look at F and F'.

To go from F to F', you need to go right in x 14 units.

Then you need to go up in y 8 units.

The translation rule is to add 14 to x and 8 to y.

(x, y) --> (x + 14, y + 8)

4 0

2 years ago

Read 2 more answers

3) What is 3x +15y=-16 in slope intercept form?

stellarik [79]

Answer:

y=-1/5x-16/15

Step-by-step explanation:

slope intercept form y=mx+b

3x+15y=-16

15y=-3x-16 divide both sides by 15

y=-3/15x-16/15 simplify -3/15 to -1/5

y=-1/5x-16/15

8 0

3 years ago

Kayla made four dresses with 7 yards of fabric how many yards did she use on one dress

lora16 [44]

1.75

Step-by-step explanation:

you want to divide 7 and 4, you should get 1.75.

5 0

3 years ago

3) Does the pair of ratios form a proportion? *<br> yes <br> no<br> please ill give brainliest⊂-:

Komok [63]

Yes it does
Explanation

8 0

3 years ago

Solve the initial value problems:<br> 1/θ(dy/dθ) = ysinθ/(y^2 + 1); subject to y(pi) = 1

ladessa [460]

Answer:

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}$

Step-by-step explanation:

Given the initial value problem $\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\$ subject to y(π) = 1. To solve this we will use the variable separable method.

Step 1: Separate the variables;

$\frac{1}{\theta}(\frac{dy}{d\theta} ) =\frac{ ysin\theta}{y^{2}+1 } \\\frac{1}{\theta}(\frac{dy}{sin\theta d\theta} ) =\frac{ y}{y^{2}+1 } \\\frac{1}{\theta}(\frac{1}{sin\theta d\theta} ) = \frac{ y}{dy(y^{2}+1 )} \\\\\theta sin\theta d\theta = \frac{ (y^{2}+1)dy}{y} \\integrating\ both \ sides\\\int\limits \theta sin\theta d\theta =\int\limits \frac{ (y^{2}+1)dy}{y} \\-\theta cos\theta - \int\limits (-cos\theta)d\theta = \int\limits ydy + \int\limits \frac{dy}{y}$

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y +C\\Given \ the\ condition\ y(\pi ) = 1\\-\pi cos\pi +sin\pi = \frac{1^{2} }{2} + ln 1 +C\\\\\pi + 0 = \frac{1}{2}+ C \\C = \pi - \frac{1}{2}$

The solution to the initial value problem will be;

$-\theta cos\thsta+sin\theta = \frac{y^{2} }{2} + ln y + \pi - \frac{1}{2}$

5 0

3 years ago