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3 years ago

To write 1.65 × 20 in standard notation, move the decimal point two places to the left. True False

Mathematics

1 answer:

Alex73 [517]3 years ago

7 0

False it would be 2 places to the RIGHT not left that would make both whole numbers which makes it standard

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a courier earns a fixed amount for each package she delivers, and yesterday her average hourly wage was $12.50 an hour. if she w

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She makes 6.25$ for each package

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3 years ago

Can you please help me graph the function using intercepts

dexar [7]

See explanation below

Explanation:

The equation: 4x + 3y = -12

To plot the graph using the intercept, we need to find the y and x intercept

To get the y intercept, we replace x with 0

4(0) + 3y = -12

0 + 3y = -12

3y = -12

divide both sides by 3:

3y/3 = -12/3

y = -4

To get the x intercept, we will replace y with 0

4x + 3(0) = -12

4x + 0 = -12

4x = -12

divide both sides by 4:

4x/4 = -12/4

x = -3

Plotting the x and y intercept on the graph:

5 0

1 year ago

A theater charges $24 for adults and $12 for senior citizens. On a recent weekends when 541 people attended the theater the tota

wel

Answer:

323 senior citizens 200 adaults

Step-by-step explanation:

7 0

3 years ago

Given: CP is perpendicular to AB, CP bisects AB, Prove: CA=CB

expeople1 [14]

Answer:

Proved CA=CB

Step-by-step explanation:

Given,

In ΔABC, CP is perpendicular to AB.

And CP bisects AB.

So, AP=PB and ∠CPA=∠CPB=90°

The figure of the triangle is in the attachment.

Now, In ΔACP and ΔBCP.

AP = PB(given)

∠CPA = ∠CPB = 90°(perpendicular)

CP = CP(common)

So, By Side-Angle-Side congruence property;

ΔACP ≅ ΔBCP

According to the property of congruence;

"If two triangles are congruent to each other then their corresponding sides are also equal."

Therefore, CA = CB (corresponding side of congruent triangle)

CA = CB Hence Proved

7 0

4 years ago

Complete the standard form of the equation of a hyperbola that has vertices at (-10, -15) and (70, -15) and one of its foci at (

Elena L [17]

Answer:

$\frac{(x-30)^{2}}{40^{2}} - \frac{(y+15)^{2}}{3^{2}} = 1$

Step-by-step explanation:

The equation of the horizontal hyperbola in standard form is:

$\frac{(x-k)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1$

The position of its center is:

$C(x,y) = \left(\frac{-10 + 70}{2}, -15 \right)$

$C(x, y) = (30,-15)$

The values for c and a are respectively:

$a = 70 - 30$

$a = 40$

$c = 30 - (-11)$

$c = 41$

The remaining variable is computed from the following Pythagorean identity:

$c ^{2} = a^{2} + b^{2}$

$b = \sqrt{c^{2}-a^{2}}$

$b = \sqrt{41^{2}-40^{2}}$

$b = 3$

Now, the equation of the hyperbola is:

$\frac{(x-30)^{2}}{40^{2}} - \frac{(y+15)^{2}}{3^{2}} = 1$

3 0

3 years ago

Read 2 more answers