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3 years ago

UMS Grade 7 Math CUA- Unit 4 / 6 of 10

Mathematics

1 answer:

brilliants [131]3 years ago

3 0

B because 48.80*0.35=17.08

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Drag and drop the symbols to enter the equation of the circle in standard form with center and radius given.

Leya [2.2K]

The standard equation of a circle with center C(a, b) and radius r is given by:

$(x-a)^2+(y-b)^2=r^2$

thus for C(a, b)=C(7, 5) and radius r=4, the equation is given by:

$(x-a)^2+(y-b)^2=r^2\\\\(x-7)^2+(y-5)^2=4^2$

Answer:

$(x-7)^2+(y-5)^2=4^2$

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4 years ago

8+3.4/2 <br> please help!!!!!

daser333 [38]

8+3 = 11.
4/2 = 1 and 1/2
Hopefully this is the stuff your looking for

8 0

4 years ago

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I need help with #5

hoa [83]

Answer:

<1 = 50

<2 = 130

Step-by-step explanation:

<1 and 130 are supplementary angles

<1 + 130 = 180

<1 =180-130

<1 = 50

130 and <2 are corresponding angles and corresponding angles are equal when the lines are parallel

<2 = 130

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3 years ago

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Please help would be very very appreciated

poizon [28]

115 ounces is the answer

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4 years ago

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Find the value of the variable y, where the sum of the fraction 2/y-3 and 6/y+3 is equal to the quotient.

NISA [10]

Answer:

Here we need to solve:

$\frac{2}{y - 3} + \frac{6}{y + 3 } = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} }$

The sum of the fractions is equal to the quotient between the fractions.

Notice that the two values:

y = 3

y = -3

make the denominator equal to zero, so those values are restricted.

We can simplify the right side to get:

$\frac{2}{y - 3} + \frac{6}{y + 3 } = \frac{\frac{2}{y-3}}{\frac{6}{y + 3} } = \frac{2*(y + 3)}{6*(y - 3)} = 3*\frac{y + 3}{y - 3}$

Now we can multiply both sides by (y - 3)

$(y - 3)*(\frac{2}{y - 3} + \frac{6}{y + 3 }) = 3*(y + 3)\\2 + 6*\frac{y -3}{y + 3} = 3*(y + 3)$

Now we can multiply both sides by (y + 3)

$(2 + 6*\frac{y -3}{y + 3})*(y + 3) = 3*(y + 3)*(y + 3)$

$2*(y + 3) + 6*(y - 3) = 3*(y + 3)*(y + 3)\\\\2*y + 6 + 6*y - 18 = 3*(y^2 + 2*y*3 + 9)\\\\8*y - 12 = 3*y^2 + 6*y + 33\\\\0 = 3*y^2 + 6*y + 33 - 8*y + 12\\\\0 = 3*y^2 - 2*y + 45$

First, let's see the determinant of that quadratic equation:

$D = (-2)^2 - 4*3*45 = -536$

We can see that it is negative, thus, there are no real solutions of the equation.

Thus, there is no value of y such that the origina equation is true,

6 0

3 years ago