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3 years ago

5

21 3/8 - 2 1/4. PLEASE TELL ME THE ANSWER

2 answers:

Elodia [21]3 years ago

7 0

Answer:

19.125 in decimals

Step-by-step explanation:

21 3/8 - 2 1/4. =19.125

Shkiper50 [21]3 years ago

4 0

Answer:

19.125

Step-by-step explanation:

21.375-2.25=19.125

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the slope is -3

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3 years ago

Which is the simplified form of m-6 p0? 1/m8p 1/m8 p/m8 m8

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3 years ago

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Find all the zeros for each function P(x)=x^4-4x^3-x^2+20x-20

ivanzaharov [21]

Answer:

The zeros of the given polynomial function are

2,2, $\pm\sqrt{5}$

Step-by-step explanation:

Given polynomial is $P(x)=x^4-4x^3-x^2+20x-20$

To find the zeros equate the given polynomial to zero

ie., P(x)=0

$P(x)=x^4-4x^3-x^2+20x-20=0$

By using synthetic division we can solve the polynomial:

2_| 1 -4 -1 20 -20

0 2 -4 -10 20

_____________________

1 -2 -5 10 |_0

Therefore x-2=0

x=2 is a zero of P(x)

Now we can write the cubic equation as below:

$x^3-2x^2-5x+10=0$

Again using synthetic division

2_| 1 -2 -5 10

0 2 0 -10

______________

1 0 -5 |_0

Therefore x-2=0

x=2 is also a zero of P(x).

Now we have $x^2+0x-5=0$

$x^2-5=0$

$x^2=5$

$x^=\pm\sqrt{5}$ is a zero of P(x)

Therefore the zeros are 2,2, $\pm\sqrt{5}$

8 0

3 years ago

Can you please answer my question number9

inn [45]

X=25... hope this helps

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3 years ago

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

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3 years ago

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