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3 years ago

The vertices of a rectangle are listed below.

Mathematics

1 answer:

elena-s [515]3 years ago

8 0

Answer:

please find attached pdf

Step-by-step explanation:

Download pdf

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4x27-3+5x2=......................

kolbaska11 [484]

Answer:

115

Step-by-step explanation:

8 0

3 years ago

wilson jarred 45 liters of jam after 5 days. how much jam did wilson jar if he spent 7 days making jam? solve using unit rates.

Natali5045456 [20]

45/5 = 9 jams a day
x/7 = 9
x= 63 jams a day

Hope this helps!

5 0

3 years ago

Why do fractions need to have a common denominator before you add or subtract them?

djyliett [7]

Fractions need common denominator to add or subtract them because without denominators a fraction is not a fraction, it is a whole number
Example - 2/8 is a fraction
but 2 is not a fraction

5 0

3 years ago

Identify the equation of the circle that has its center at (16, 30) and passes through the origin

Veronika [31]

To solve this question, we have to find the equation of the circle with given center and where it passes. Doing this, we get that the equation of the circle is:

$(x - 16)^2 + (y - 30)^2 = 1156$

Equation of a circle:

The equation of a circle with center $(x_0, y_0)$ and radius r is given by:

$(x - x_0)^2 + (y - y_0)^2 = r^2$

Center at (16, 30)

This means that $x_0 = 16, y_0 = 30$

Thus

$(x - 16)^2 + (y - 30)^2 = r^2$

Passes through the origin:

We use this to find the radius squared, as this means that $x = 0, y = 0$ is part of the circle. Thus

$(x - 16)^2 + (y - 30)^2 = r^2$

$(0 - 16)^2 + (0 - 30)^2 = r^2$

$r^2 = 16^2 + 30^2 = 1156$

Thus, the equation of the circle is:

$(x - 16)^2 + (y - 30)^2 = 1156$

For another example to find the equation of a circle, you can look at brainly.com/question/23719612

4 0

3 years ago

Use mathematical induction to show that 4^n ≡ 3n+1 (mod 9) for all n equal to or greater than 0

cestrela7 [59]

When $n=0$ , you have

$4^0=1\equiv3(0)+1=1\mod9$

Now assume this is true for $n=k$ , i.e.

$4^k\equiv3k+1\mod9$

and under this hypothesis show that it's also true for $n=k+1$ . You have

$4^k\equiv3k+1\mod9$
$4\equiv4\mod9$
$\implies 4\times4^k\equiv4(3k+1)\mod9$
$\implies 4^{k+1}\equiv12k+4\mod9$

In other words, there exists $M$ such that

$4^{k+1}=9M+12k+4$

Rewriting, you have

$4^{k+1}=9M+9k+3k+4$
$4^{k+1}=9(M+k)+3k+3+1$
$4^{k+1}=9(M+k)+3(k+1)+1$

and this is equivalent to $3(k+1)+1$ modulo 9, as desired.

3 0

3 years ago