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3 years ago

Tiana’s car traveled 111 miles on 3/8 of a tank of gas. How far will she be able to go on a full tank of gas?

Mathematics

1 answer:

Anon25 [30]3 years ago

4 0

Answer:

Hiiiiiii

Step-by-step explanation:

So to get the answer you have to turn the 3/8 into a mixed number and divide it by 111

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How to graph y <7/3x- 5

fenix001 [56]

Use Desmos the graphing app it's great!

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3 years ago

If the original square had a side length of

irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$

$A2=(4)(x)=4x\ in^2$

$A3=(x)(7)=7x\ in^2$

$A4=(4)(7)=28\ in^2$

The total area of the second rectangle is the sum of the four areas

$A=A1+A2+A3+A4$

State the product of (x+4) and (x+7) as a trinomial

$(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28$

Part c) If the original square had a side length of x = 2 inches, then what is the area of the second rectangle?

we know that

The area of the second rectangle is equal to

$A=A1+A2+A3+A4$

For x=2 in

substitute the value of x in the area of each portion

$A1=(2)(2)=4\ in^2$

$A2=(4)(2)=8\ in^2$

$A3=(2)(7)=14\ in^2$

$A4=(4)(7)=28\ in^2$

$A=4+8+14+28$

$A=54\ in^2$

Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 in

We have that

The trinomial is

$A(x)=x^{2} +11x+28$

For x=2 in

substitute and solve for A(x)

$A(2)=2^{2} +11(2)+28$

$A(2)=4 +22+28$

$A(2)=54\ in^2$ ----> verified

therefore

The trinomial represent the total area of the second rectangle

7 0

3 years ago

62+x=82 Does any one know this answer

Vladimir [108]

62 + x = 82

Subtract both sides by 62.

x = 20.

4 0

3 years ago

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Find the sine of angle S. Reduce the answer to the lowest terms.

NeX [460]

Answer:

$\sin e\text{ S = }\frac{3}{5}$

Explanation:

Here, we want to get the value of sine S

The sine of an angle is the ratio of the opposite side to the hypotenuse side

The hypotenuse side is the side that faces the right angle which is the small box. It is the longest side of the triangle and it measures 10

The opposite is the side that faces the angle which is 6

We have the sine as:

$\frac{6}{10}\text{ = }\frac{3}{5}$

8 0

11 months ago

Which statement proves that quadrilateral JKLM is a kite?

sesenic [268]

A quadrilateral is a kite if the diagonals are:

i) perpendicular
ii) bisect each other
iii) not equal ( together with conditions i and ii this would make the quadrilateral a square)

Another definition of the kite is :

a quadrilateral with 2 pairs of equal adjacent sides.

Let's check the choices one by one:

A. ∠M is a right angle and MK bisects ∠LMJ.

according to these, ML and MJ may well be not equal...

B. LM = JM = 3 and JK = LK = √17.

this makes the quadrilateral a kite.

C. MK intersects LJ at its midpoint

if they are not perpendicular, the quadrilateral is not a kite.

D. The slope of MK is –1 and the slope of LJ is 1.

this only means that MK and LJ are perpendicular, but not whether they bisect each other,

Answer: only B

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3 years ago

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