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2 years ago

A 15- ft tree casts an 18- ft shadow at the same time a 24- ft tree casts a shadow. How long is the shadow of the 24- ft tree

Mathematics

1 answer:

anzhelika [568]2 years ago

4 0

Answer:

28.8 ft

Step-by-step explanation:

We can use proportion in calculating this.

Let X= height of the shadow of 24- ft tree.

The proportion can be set up as

(15- ft tree) /( 18- ft shadow )=(24- ft tree) / X

15/18 = 24/X

Making X subject of the formula

15X = 18 × 24

X= (18×24)/15

X=28.8 ft

Hence, the shadow of the 24- ft tree is 28.8 feet

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fiasKO [112]

The diagram shows that FG = IJ are congruent segments due to the double tickmarks. These are the hypotenuses of the right triangles.

So that takes care of the "H" in "HL".

The L stands for "leg", so we need to have a congruent pair of corresponding legs.

Due to the orientation of the triangles, it's not clear how the corresponding legs match up, but if any of the following is true

EF = IK
EF = JK
EG = IK
EG = JK

then we'd have enough to use HL.

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2 years ago

The number of people who live in a unit of area is called the population density of the area. It usually given as people "per sq

bazaltina [42]

Answer: 0.03

Step-by-step explanation:

1 block = 1/20 mile = 0.05 miles

4(0.05) = 0.20

3(0.05) = 0.15

A=l•w=0.20(0.15)=0.03

8 0

2 years ago

If ΔOPQ is dilated from point Q by a scale factor of 3, which of the following equations is true about segment R prime S prime?

Sergio [31]

Answer:

$\overline {R'S'} = 3 {\overline {RS} }$

Step-by-step explanation:

The scale factor of dilation of triangle ΔOPQ, S.F. = 3

The center of dilation = Point <em>Q</em>

Therefore;

The length of the segment, ${\overline {S'Q'} }$ = 3 × The length of the segment, ${\overline {SQ} }$

Similarly;

The length of the segment, ${\overline {R'Q'} }$ = 3 × The length of the segment, ${\overline {RQ} }$

By Pythagoras theorem, we have;

${\overline {R'Q'} }^2 = {\overline {R'S'} }^2 + {\overline {S'Q'} }^2$

Therefore;

${\overline {R'S'} }^2 = {\overline {R'Q'} }^2 - {\overline {S'Q'} }^2 = \left ({3 \times \overline {RQ} } \right) ^2 - \left (3 \times {\overline {SQ} } \right) ^2 = 9 \times \left (\overline {RQ} } ^2 - {\overline {SQ} } ^2 \right) = 9 \times {\overline {RS} }^2$

$\therefore \sqrt{ {\overline {R'S'} }^2} = \sqrt{ 9 \times {\overline {RS} }^2}= 3 \times {\overline {RS} }$

$\overline {R'S'} = 3 \times {\overline {RS} }$ .

6 0

2 years ago

Allushta [10]

Answer:

Step-by-step explanation:

We can use Hypotenuse Theorem for this:

For triangle ABC,

AC²=CB²+BA²

9²=CB²+3²

Switch the sides

-CB²=3² -9²

-CB²= 9-81

-CB²= -72

the signs cut each other

CB²=72

the square changes sides

CB=√72

CB= 8.49 cm

Now triangle CBD

We have the length of CB but not the other two whereas we have the angle

It will be sin

As we know sin= O/H

Here O is CB whereas H is CD

Put the values

sin 50= 8.49/H

H= 8.49 ÷sin 50

Calculate

H=11.1 cm

As we know H is CD so CD is 11.1 cm

Hope it helps

8 0

2 years ago

Gerry wants to construct a triangle with segments that

alexira [117]

Answer:

See Explanation

Step-by-step explanation:

Sample Response: No, the triangle is not possible. The triangle inequality rule states the sum of any two sides must be greater than the third side. 6 + 4 is 10, which is less than 11.

5 0

2 years ago