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2 years ago

5

What is a decimal in mathematics?

1 answer:

inn [45]2 years ago

5 0

A decimal is 3.5, or 2.4739464. a number and a little bit. basically not a whole number. :)

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Point H is located at (2,−7)

Mekhanik [1.2K]

Answer:

x axis & (-5,-7)

Step-by-step explanation:

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2 years ago

WILL ANSWER FIRST RIGHT ANSWER AS BRAINIEST

Ket [755]

Answer:

8

Step-by-step explanation:

Considering the given value of sum of interior angles with exclusion of one to be 1070 then we can write it as

1070=180(n-2) where n is the number of sides required here.

Making n the subject of the formula then following step by step

$n-2=\frac {1070}{180}$

$n=\frac {1070}{180}+2\\n=7.9444444444444\approx 8$

Therefore, the shape is octagon and it has 8 sides.

8 0

2 years ago

What is the quotient of StartFraction 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline EndFraction? Star

Komok [63]

The quotient of the number given number 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline is 7.

<h3>What is the quotient?</h3>

Quotient is the resultant number which is obtain by dividing a number with another. Let a number a is divided by number b. Then the quotient of these two number will be,

$q=\dfrac{a}{b}$

Here, (<em>a, b</em>) are the real numbers.

The number StartFraction 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline EndFraction, given can be written as,

$\dfrac{7^{-1}}{7^{-2}}$

Let the quotient of this division is n. Therefore,

$n=\dfrac{7^{-1}}{7^{-2}}$

A number in numerator of a fraction with negative exponent can be written in the denominator with the same but positive exponent and vise versa. Therefore,

$n=\dfrac{7^{2}}{7^{1}}\\n=7$

Hence, the quotient of the number given number 7 Superscript negative 1 Baseline Over 7 Superscript negative 2 Baseline is 7.

Learn more about the quotient here;

brainly.com/question/673545

7 0

2 years ago

How many time can 25 go in 389

iren2701 [21]

15 times is the answer

5 0

2 years ago

Read 2 more answers

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

2 years ago

Read 2 more answers