Could you guys please help me solve this basic math. PLS OMG ITS DUE :(

A local hamburger shop sold a combined total of 763 hamburgers and cheeseburgers on Saturday. There were 63 more cheeseburgers s

Airida [17]

In order to solve this we'll start by assigning variables to hamburgers and cheeseburgers, since these are what we're trying to find. Lets say x = hamburgers and y = cheeseburgers. So we know two things, we know that x+y= 763 (hamburgers plus cheeseburgers sold equals 763, and we know that y= x+63 (cheeseburgers sold equals 63 more than hamburgers sold). Now we have a system of equations. This can be solved most easily by rearranging each equation to each y, and then set them equal to each other:
x+y=763 -> y=763-x, and we already have y=x+63. Set them equal to each other:
x+63 = 763-x (add x to both sides) -> 2x+63 = 763 (subtract 63 from both sides) -> 2x = 700 (divide both sides by 2) x = 350. So we solved for x, which is hamburgers sold, which is what the question asks for, so your answer is 350 hamburgers were sold on Saturday

6 0

3 years ago

In AOPQ, o = 700 cm, p = 840 cm and q=620 cm. Find the measure of _P to the nearest degree

Westkost [7]

Given:

In triangle OPQ, o = 700 cm, p = 840 cm and q=620 cm.

To find:

The measure of angle P.

Solution:

According to the Law of Cosines:

$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$

Using Law of Cosines in triangle OPQ, we get

$\cos P=\dfrac{o^2+q^2-p^2}{2oq}$

$\cos P=\dfrac{(700)^2+(620)^2-(840)^2}{2(700)(620)}$

$\cos P=\dfrac{490000+384400-705600}{868000}$

$\cos P=\dfrac{168800}{868000}$

On further simplification, we get

$\cos P=0.19447$

$P=\cos^{-1}(0.19447)$

$P=78.786236$

$P\approx 79$

Therefore, the measure of angle P is 79 degrees.

8 0

3 years ago

Read 2 more answers

Tell which property the statement x+14=14+x illustrates.

Naddik [55]

Answer:associative property of addition

Step-by-step explanation:The associative property of addition says that changing the grouping of the addends does not change the sum.

8 0

2 years ago

(a) Find a vector parallel to the line of intersection of the planes −4x+2y−z=1 and 3x−2y+2z=1.

valentinak56 [21]

Find the intersection of the two planes. Do this by solving for z in terms of x and y ; then solve for y in terms of x ; then again for z but only in terms of x.

-4x + 2y - z = 1 ==> z = -4x + 2y - 1

3x - 2y + 2z = 1 ==> z = (1 - 3x + 2y)/2

==> -4x + 2y - 1 = (1 - 3x + 2y)/2

==> -8x + 4y - 2 = 1 - 3x + 2y

==> -5x + 2y = 3

==> y = (3 + 5x)/2

==> z = -4x + 2 (3 + 5x)/2 - 1 = x + 2

So if we take x = t, the line of intersection is parameterized by

r(t) = ⟨t, (3 + 5t )/2, 2 + t⟩

Just to not have to work with fractions, scale this by a factor of 2, so that

r(t) = ⟨2t, 3 + 5t, 4 + 2t⟩

(a) The tangent vector to r(t) is parallel to this line, so you can use

v = dr/dt = d/dt ⟨2t, 3 + 5t, 4 + 2t⟩ = ⟨2, 5, 2⟩

or any scalar multiple of this.

(b) (-1, -1, 1) indeed lies in both planes. Plug in x = -1, y = 1, and z = 1 to both plane equations to see this for yourself. We already found the parameterization for the intersection,

r(t) = ⟨2t, 3 + 5t, 4 + 2t⟩