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zubka84 [21]
4 years ago
14

Please helppppppppppppppppppp

Mathematics
2 answers:
Xelga [282]4 years ago
5 0
First one is 10/14 and 15/21
Second answer is 16/18 and 24/27
Give thanks!
Gennadij [26K]4 years ago
4 0
Multiply each by the same constant and you'll get another ratio ie. 5/7 --> 10/14 --> 15/21
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Solve the equation.<br> -13 = 2b - b - 10
astra-53 [7]

Answer:

b = -3

Step-by-step explanation:

-13 = 2b - b - 10

Add 10 to both sides of the equation

-3 = 2b - b

Subtract 2b - b

-3 = b

b = -3

Hope this helps :)

6 0
4 years ago
How.many grams equals 25 killograms
34kurt

Answer:

Step-by-step explanation:

1 kilogram = 1000 grams

since we need to know 25 kilograms you can do

25 x 1000

giving us 25,000

4 0
3 years ago
Read 2 more answers
What is the length of segment AB?
allochka39001 [22]
The answer to this question is 8
4 0
4 years ago
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if 1 000 bottles of an award-winning wine are to be packed in special gift boxes which can hold six bottles each, calculate how
Anuta_ua [19.1K]

Step-by-step explanation:

Introduction:

This section basically dedicated to the classification of the antennas which are used in different wavelength.

Wire Antennas:

Wire antennas are familiar to the layman because they are seen virtually everywhere on automobiles, buildings, ships, aircraft, spacecraft, and so on.

There are various shapes of wire antennas such as a straight wire (dipole), loop, and helix which are shown in Figure 1.3.

Loop antennas need not only be circular.

They may take the form of a rectangle, square, ellipse, or any other configuration.

The circular loop is the most common because of its simplicity in construction

Aperture Antennas

Aperture antennas may be more familiar to the layman today than in the past because of the increasing demand for more sophisticated forms of antennas and the utilization of higher frequencies.

Some forms of aperture antennas are shown in Figure 1.4.

Antennas of this type are very useful for aircraft and spacecraft applications, because they can be very conveniently flush-mounted on the skin of the aircraft or spacecraft. In addition, they can be covered with a dielectric material to protect them from hazardous conditions of the environment.

Microstrip Antennas

Microstrip antennas became very popular in the 1970s primarily for spaceborne applications.

Today they are used for government and commercial applications. These antennas consist of a metallic patch on a grounded substrate.

The microstrip antennas are low profile, comformable to planar and nonplanar surfaces, simple and inexpensive to fabricate using modern printed-circuit technology, mechanically robust when mounted on rigid surfaces, compatible with MMIC designs, and very versatile in terms of resonant frequency, polarization, pattern, and impedance.

These antennas can be mounted on the surface of high-performance aircraft, spacecraft, satellites, missiles, cars, and even handheld mobile telephones

Array Antennas

Many applications require radiation characteristics that may not be achievable by a single element. It may, however, be possible that an aggregate of radiating elements

in an electrical and geometrical arrangement (an array) will result inthe desiredradiation characteristics.

The arrangement of the array may be such that the radiation from the elements adds up to give a radiationmaximum ina particular directionor directions, minimum in others, or otherwise as desired.

Typical examples of arrays are shownin Figure 1.6.

Usually the term array is reserved for an arrangement in which the individual radiators are separate as shown in Figures 1.6(a/c).

However the same term is also used to describe an assembly of radiators mounted on a continuous structure, showninFigure 1.6(d).

6 0
2 years ago
Read 2 more answers
=) The perimeter of a rectangular field is 336 m.
goldenfox [79]

Answer:

Length of the field: 94 m

Width of the painting: 61 cm

Step-by-step explanation:

Use the perimeter formula, P = 2l + 2w, to find the length:

Plug in the perimeter and width into the equation:

P = 2l + 2w

336 = 2l + 2(74)

336 = 2l + 148

188 = 2l

94 = l

So, the length of the field is 94 m.

To find the width of the painting, use the area formula, A = lw

Plug in the area and length into the equation:

A = lw

5795 = 95w

61 = w

So, the width of the painting is 61 cm.

Length of the field: 94 m

Width of the painting: 61 cm

4 0
3 years ago
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