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3 years ago

14

-6-4+[-4] 0 [-4-[-3]] [4² + 3²] ÷ 5

1 answer:

mestny [16]3 years ago

7 0

Answer:

you can eat this panipuri then you will got answer

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What is the equation in slope intercept form of the line that passes through the point(5,0)and is parallel to the line represent

Masteriza [31]

Answer:

Step-by-step explanation:

y - 0 = 1.2(x - 5)

y = 1.2x - 6

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3 years ago

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Write the equation in slope intercept form

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Your answer is

slope-intercept equation is y=-3x+6

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3 years ago

A deli receives a large order for sandwiches for a company picnic. The company wants one club sandwich and two regular sandwiche

lions [1.4K]

15 because it is 7 per person because 2 regulars and 1 club, and then they want the one Dagwood which is 5. So overall it is (110-5)/7 which equals 105/7 which is 15

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3 years ago

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How to solve this plss frac{(\sqrt{(3)})^{5}}{((\sqrt{(3)})^{-4})}=(\sqrt{(3)})^{(2k+1)

dusya [7]

Answer:

k = 4

Step-by-step explanation:

Given equation:

$\dfrac{(\sqrt{3})^{5}}{(\sqrt{3})^{-4}}=(\sqrt{3})^{(2k+1)}$

$\textsf{Apply exponent rule} \quad \dfrac{a^b}{a^c}=a^{b-c}:$

$\implies (\sqrt{3})^{(5-(-4))}=(\sqrt{3})^{(2k+1)}$

$\implies (\sqrt{3})^{9}=(\sqrt{3})^{(2k+1)}$

$\textsf{Apply exponent rule} \quad a^{f(x)}=a^{g(x)} \implies f(x)=g(x):$

$\implies (\sqrt{3})^{9}=(\sqrt{3})^{(2k+1)}$

$\implies 9=2k+1$

$\implies 2k=8$

$\implies k=4$

6 0

2 years ago

A circle is centered at the point (-3, 2) and passes through the point (1, 5). The radius of the circle is ___ units. The point

mixas84 [53]

Answer:

The radius is 5 units

(-7,-1) and (-7,5) lies on the circle

Step-by-step explanation:

Use the distance formula to find the distance between the center and the given point that gives you the radius.

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

The center is (-3,2) and the point on the circumference is (1,5)

$r=\sqrt{(1--3)^2+(5-2)^2}$

$r=\sqrt{(4)^2+(3)^2}$

$r=\sqrt{16+9}$

$r=\sqrt{25}$

$\therefore r=5$

The radius is 5 units.

The equation of the circle is given by:

$(x--3)^2+(y-2)^2=5^2$

$(x+3)^2+(y-2)^2=25$

When x=-7, we obtain y by substituting into the equation of the circle:

$(-7+3)^2+(y-2)^2=25$

$(-4)^2+(y-2)^2=25$

$16+(y-2)^2=25$

$(y-2)^2=25-16$

$(y-2)^2=9$

$y-2=\pm \sqrt{9}$

$y=2\pm 3$

$y=-1$ or y=5

Therefore (-7,-1) and (-7,5) lies on the circle.

6 0

4 years ago