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3 years ago

Estimate -9 1/6+2 1/3+(-1 7/8)

Mathematics

1 answer:

Eduardwww [97]3 years ago

5 0

Answer:

-9

Step-by-step explanation:

-9+2-2=-9

You might be interested in

A. What are the coordinates of the vertices of the triangle below?

Viktor [21]

Answer:

Solution given:

A.coordinate are

A(-2,3)

B(0,-3)

C(4,5)

B.Each length are :

we have

length $\sqrt{(x2-x1)²+(y2-y1)²}$

now

AB: $\sqrt{(-2-0)²+(3+3)²}$ = $2\sqrt{10}$ units

BC: $\sqrt{(0-4)²+(-3-5)²}$ = $4\sqrt{5}$ units

AC: $\sqrt{(-2-4)²+(3-5)²}$ = $2\sqrt{10}$ units

C. the figure:

By using Pythagoras law

base[b]=AB=perpendicular [p]=AC

hypotenuse [h]=BC

we have

h²=p²+b²

substituting value

( $4\sqrt{5}$ )²=2p²

16*5=2*( $2\sqrt{10}$ )²

80=2*4*10

80=80

SO IT IS RIGHT ANGLED ISOSCELES TRIANGLE.

4 0

3 years ago

What is the solution to the system of equations?

castortr0y [4]

Consider this option/solution.
P.S. The method of solution is Gauss' method.

3 0

3 years ago

Find the domain and range (0,1), (3,1), (3,9)

Elza [17]

Answer:

domain={0,3}

range={1,9}

Step-by-step explanation:

7 0

3 years ago

Help needed, much appreciated

mrs_skeptik [129]

Answer:

1/6

Explain:

The reciprocal of the slope is its perpendicular.

6/1 --> 1/6

3 0

3 years ago

Find the perimeter P of ▱JKLM with vertices J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2). Round your answer to the nearest tenth, if

Bezzdna [24]

Given:

Vertices of JKLM are J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2).

To find:

The perimeter P of a parallelogram JKLM.

Solution:

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$JK=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(-5-\left(-2\right)\right)^2}$

$JK=\sqrt{\left(-5+3\right)^2+\left(-5+2\right)^2}$

$JK=\sqrt{\left(-2\right)^2+\left(-3\right)^2}$

$JK=\sqrt{4+9}$

$JK=\sqrt{13}$

Similarly,

$KL=\sqrt{\left(1-\left(-5\right)\right)^2+\left(-5-\left(-5\right)\right)^2}=6$

$LM=\sqrt{\left(3-1\right)^2+\left(-2-\left(-5\right)\right)^2}=\sqrt{13}$

$JM=\sqrt{\left(3-\left(-3\right)\right)^2+\left(-2-\left(-2\right)\right)^2}=6$

Now, perimeter P of ▱JKLM is

$P=JK+KL+LM+JM$

$P=\sqrt{13}+6+\sqrt{13}+6$

$P=2\sqrt{13}+12$

$P=2(3.61)+12$

$P=7.22+12$

$P=19.22$

$P\approx 19.2$

Therefore, the perimeter P of ▱JKLM is 19.2 units.

3 0

3 years ago