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s344n2d4d5 [400]

3 years ago

14

What is 1+9-8+40-7-8+100-20

2 answers:

natka813 [3]3 years ago

5 0

Answer: 107

I did the math on my calculator, just put it in the calculator, like it is, it follows PEMDAS, the way its written.

luda_lava [24]3 years ago

4 0

Answer:

$107$

Step-by-step explanation:

$1+9-8+40-7-8+100-20$

$=10-8+40-7-8+100-20$

$=2+40-7-8+100-20$

$=42-7-8+100-20$

$=35-8+100-20$

$=27+100-20$

$=127-20$

$=107$

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Write the fraction or mixed number as a money amount and as a decimal in terms of dollars. 1 48/100

Kaylis [27]

Answer:

1.2.4.3.4

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

What are the two ordered pairs for 5x=3

valkas [14]

Well the equation of this line is $x=3/5$ . Which means that for any value of y the x will always equal to 3/5, so the points (ordered pairs) for eg. are $(3/5,2),(3/5,\sqrt{7})$ .

Hope this helps.

3 0

3 years ago

The coordinates of P and Q are (2,6) and (4,5) respectively. Equation of a line RS is Y= 2X -3. Determine if the lines are paral

Travka [436]

Answer:

Perpendicular

Step-by-step explanation:

Given points P=(2,6) and Q=(4,5)

The line RS is y=2x-3.

Now we know that the line joining any two points (x1,y1) and (x2,y2) has slope m= $\frac{y2-y1}{x2-x1}$

Now the slope of the line PQ is

m1= $\frac{5-6}{4-2}$ = - $\frac{1}{2}$

We know that the slope of the line in the form y=mx+c is m

So the line RS has slope m2=2

Now two lines are parallel if m1=m2

And those two lines are perpendicular if $m1*m2=-1$

here m1= - $\frac{1}{2}$ and m2=2

clearly $m1*m2=-1$

Therefore the two given lines are perpendicular

7 0

3 years ago

Which point lies on a circle with a radius of 5 units and center at P(6,1)?

ryzh [129]

Answer:

Option B. $R(2,4)$

Step-by-step explanation:

we know that

If a ordered pair lie on the circle. then the ordered pair must satisfy the equation of the circle

step 1

Find the equation of the circle

we know that

The equation of the circle in center radius form is equal to

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

where

r is the radius of the circle

(h,k) is the center of the circle

substitute the values

$(x-6)^{2}+(y-1)^{2}=5^{2}$

$(x-6)^{2}+(y-1)^{2}=25$

step 2

Verify each case

case A) $Q(1, 11)$

substitute the value of $x=1, y=11$ in the equation of the circle and then compare the results

$(1-6)^{2}+(11-1)^{2}=25$

$25+100=25$ ------> is not true

therefore

the ordered pair Q not lie on the circle

case B) $R(2,4)$

substitute the value of $x=2, y=4$ in the equation of the circle and then compare the results

$(2-6)^{2}+(4-1)^{2}=25$

$16+9=25$ ------> is true

therefore

the ordered pair R lie on the circle

case C) $S(4,-4)$

substitute the value of $x=4, y=-4$ in the equation of the circle and then compare the results

$(4-6)^{2}+(-4-1)^{2}=25$

$4+25=25$ ------> is not true

therefore

the ordered pair S not lie on the circle

case D) $T(9,-2)$

substitute the value of $x=4, y=-4$ in the equation of the circle and then compare the results

$(9-6)^{2}+(-2-1)^{2}=25$

$9+9=25$ ------> is not true

therefore

the ordered pair T not lie on the circle

7 0

4 years ago

Pls help i am on a timer

stich3 [128]

Answer:

$x = 1 + \sqrt{2} \ \ or \ 1- \sqrt{2}$

Step-by-step explanation:

Given;

x² - 2x - 1 = 0

Solve by completing the square method;

⇒ take the constant to the right hand side of the equation.

x² - 2x = 1

⇒ take half of coefficient of x = ¹/₂ x -2 = -1

⇒ square half of coefficient of x and add it to the both sides of the equation

$x^2 + (-1)^2 = 1 + (-1)^2$

$(x-1)^2 = 1 + 1\\\\(x-1)^2 = 2\\\\$

⇒ take the square root of both sides;

$x-1 = +/- \ \ \sqrt{2} \\\\x = 1 + \sqrt{2} \ \ or \ 1- \sqrt{2}$

Therefore, option B is the right solution.

3 0

3 years ago