Ask question

Ask question

All categories

3 years ago

11

I need your Help, please! THX:)

2 answers:

USPshnik [31]3 years ago

7 0

450 miles
1 in. Per 60 miles
460/60 = 7.5 inches

Allushta [10]3 years ago

5 0

Answer:

7.5 inches

Step-by-step explanation:

We have the scale factor 1:60.

So we can setup the equation: 1/60=x/450.

Simplify denominators: 1/2 = x/15

Cross multiply: 2x=15, so x=7.5

You might be interested in

If JKLM is a rhombus, MK = 30, NL = 13, and mZMKL = 41°, find each measure.

oksian1 [2.3K]

Answer:

$NK = 15$

$JL = 26$

$KL = 19.85$

$\angle JKM =49$

$\angle JML =41$

$\angle MLK = 90$

$\angle MNL =90$

$\angle KJL =41$

Step-by-step explanation:

Given

$MK = 30$

$NL = 13$

$\angle MKL = 41$

Solving (a): NK

MK is a diagonal and NK is half of the diagonal. So:

$NK = \frac{1}{2} * MK$

$NK = \frac{1}{2} * 30$

$NK = 15$

Solving (b): JL

JL is a diagonal, and it is twice of NL.

$JL = 2 * NL$

$JL = 2 * 13$

$JL = 26$

Solving (c): KL

To solve for KL, we consider triangle KNL where:

$\angle KNL = 90$

and

$KL^2 = NL^2 + NK^2$

$KL^2 = 13^2 + 15^2$

$KL^2 = 394$

$KL = \sqrt{394$

$KL = 19.85$

Solving (d - h):

To do this, we consider triangle JKN

$\angle KNL = \angle LNM = \angle MNJ = \angle JNK = 90$ -- diagonals bisect one another at right angle

Alternate interior angles are equal. So:

$\angle MKL = \angle KMJ = \angle KJL = \angle JLM = 41$

Similarly:

$\angle MKJ = \angle KML = \angle MJL = \angle JLK = 90 - 41$

$\angle MKJ = \angle KML = \angle MJL = \angle JLK = 49$

So:

$\angle JKM =49$

$\angle JML =41$

$\angle MLK = \angle MLJ + \angle JLK$

$\angle MLK = 49 + 41$

$\angle MLK = 90$

$\angle MNL =90$

$\angle KJL =41$

5 0

3 years ago

The temperature was -9°F in the morning and rose to a temperature of 17°F in the afternoon. What was the change in temperature f

Anastasy [175]

The change in temperature for the day was 26 degrees F

5 0

3 years ago

. Solve the given equation <br>3m+7<br>____=13/6<br>5m-4

oee [108]

Answer is X=2
Merry Christmas

5 0

3 years ago

Read 2 more answers

I need help asap.

skad [1K]

Answer:

435=107+82+246

Step-by-step explanation:

I think this is correct. Someone correct me if I am wrong, but I took 435 and subtracted 107, then I took that number and divided it by 4 and took one of those 4 and put it in the equation then multiplied the last three.

3 0

2 years ago

X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

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

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

4 0

2 years ago