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statuscvo [17]
3 years ago
11

I need your Help, please! THX:)

Mathematics
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

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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
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