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3 years ago

Pls help meeeee thank you

Mathematics

1 answer:

Alex787 [66]3 years ago

6 0

Answer:

Step-by-step explanation: learn yourself yw -dabay

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Are the following fractions proportional?? .4/.8 and .71/.4

Neporo4naja [7]

Answer:

yes they are

Step-by-step explanation:

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3 years ago

Is a diameter of 9cm more than a radius of 5cm and which would have the larger circumference?

Tpy6a [65]

Answer:

Circle B

3.14 cm

Step-by-step explanation:

1. The diameter is twice the radius, so the diameter of Circle B is ...

2 × 5 cm = 10 cm

This diameter is greater than the 9 cm diameter of Circle A, so Circle B is larger and will have a larger circumference.

2. The circumference is the product of π and the diameter. Then the difference between the two circumferences is ...

... (circle B circumference) - (circle A circumference) = π·(10 cm) - π·(9 cm)

... = π·(10 -9) cm = π·1 cm ≈ 3.14 cm

6 0

2 years ago

4x - 6 + x + 3<br><br> ASAP BRAINLIEST

Pachacha [2.7K]

Answer:

5x - 3

Step-by-step explanation:

Add like terms: (4x + x) + (-6 + 3) = 5x - 3

3 0

3 years ago

Read 2 more answers

find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

6 0

2 years ago

Brandon is shopping at Old Navy during a sale. All shorts are $16 each and all t-shirts are $10 each.He has $100 to spend and wo

DaniilM [7]

Let
x = the number of shorts bought
y = the number of t-shirts bought

A pair of shorts costs $16 and a t-shirt costs $10. Brandom has $100 to spend.
Therefore
16x + 10y ≤ 100
This may be written as
y ≤ - 1.6x + 10 (1)

Brandon wants at least 2 pairs of shorts. Therefore
x ≥ 2 (2)

Graph the equations y = -1.6x + 10 and x = 2.
The shaded region satisfies both inequalities.

Answer:
Two possible solutions are
(a) 3 pairs of shorts and 4 t-shirts,
(b) 4 pairs of shorts and 2 t-shirts.

7 0

2 years ago