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Oxana [17]
3 years ago
5

ASAP Mario was given the following enlargement. Side A corresponds to a side with length 24 centimeters. A side with length 6 ce

ntimeters corresponds with a side with length 12 centimeters. Figures not drawn to scale. What is the length of side A, in centimeters? 12 15 24 33
Mathematics
2 answers:
german3 years ago
4 0

Answer:

ANSWER:A

Step-by-step explanation:

ANSWER IS A!SAVE YOURSELF! RUN! DON'T FAIL UR TEST!

byeeeeee:)

<3

Usimov [2.4K]3 years ago
3 0

Answer:

The length of side A is 12 cm

Step-by-step explanation:

This question is best answered when there's an attachment; however, the absence of an attachment doesn't mean it can't be solved.

Given

Side A corresponds to 24 cm

Length 6 cm corresponds to Length 12 cm

Required

Find A

The given parameters can be represented mathematically as;

\frac{A}{24} = \frac{6}{12}

With this expression, the value of A can be easily solved

First; multiply both sides by 24

24 * \frac{A}{24} = \frac{6}{12} * 24

\frac{24 *A}{24} = \frac{6}{12} * 24

\frac{24A}{24} = \frac{6}{12} * 24

A = \frac{6}{12} * 24

A = \frac{6 * 24}{12}

A = \frac{144}{12}

A =12

<em>Hence, the length of side A is 12 cm</em>

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DENIUS [597]

Answer:

a. ∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))

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d. ∀x (ThroughputNormal ∧(ProxyServer(x)∧ ¬Diagnostic(x))) → (∃y Router(y)∧Functioning(y))

Step-by-step explanation:

a)  

Let the domain be users and mailboxes. Let User(x) be “x is a user”, let Mailbox(y) be “y is a mailbox”, and let Access(x, y) be “x has access to y”.  

∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))  

(b)

Let the domain be people in the group. Let Access(x, y) be “x has access to y”. Let FileSystemLocked be the proposition “the file system is locked.” Let System Mailbox be the constant that is the system mailbox.  

FileSystemLocked → ∀x Access(x, SystemMailbox)  

(c)  

Let the domain be all applications. Let Firewall(x) be “x is the firewall”, and let ProxyServer(x) be “x is the proxy server.” Let Diagnostic(x) be “x is in a diagnostic state”.  

∀x ∀y ((Firewall(x) ∧ Diagnostic(x)) → (ProxyServer(y) → Diagnostic(y))  

(d)

Let the domain be all applications and routers. Let Router(x) be “x is a router”, and let ProxyServer(x) be “x is the proxy server.” Let Diagnostic(x) be “x is in a diagnostic state”. Let ThroughputNormal be “the throughput is between 100kbps and 500 kbps”. Let Functioning(y) be “y is functioning normally”.  

∀x (ThroughputNormal ∧(ProxyServer(x)∧ ¬Diagnostic(x))) → (∃y Router(y)∧Functioning(y))

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3 years ago
Suppose that 45% of people have dogs. If two people are randomly chosen, what is the probability that they both have a dog
mylen [45]

Answer:

P(Dogs) = 0.2025

Step-by-step explanation:

Given

Proportion, p = 45\%

Required

Probability of two people having dog

First, we have to convert the given parameter to decimal

p = \frac{45}{100}

p = 0.45

Let P(Dogs) represent the required probability;

This is calculated as thus;

<em>P(Dogs) = Probability of first person having a dog * Probability of second person having a dog</em>

<em />

P(Dogs) = p * p

P(Dogs) = 0.45 * 0.45

P(Dogs) = 0.45^2

P(Dogs) = 0.2025

<em>Hence, the probability of 2 people having a dog is </em>P(Dogs) = 0.2025<em />

4 0
3 years ago
Please help thank you
ioda
<h2>Here </h2>

Tan65°=<u>h/</u><u>7</u><u>5</u><u>m</u>

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Step-by-step explanation:

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17. The Bricklayer needs 100 pieces of angle, each to be cut 54 3/16" long. 1 point
mariarad [96]

The number of cuts from a 20% length is 2 and the number of 20' angle are needed in order to have the 100 pieces is 5

<h3>How to determine the number</h3>

From the information given;

The bricklayer needs = 100 pieces of angle

The number of cuts = 54

Length of cuts = 3/ 16'' Inches

a. To determine the number of cuts form a 20% length, we have to multiply the length by the percentage and by the initial number of cuts

= 20/ 100 × 3/ 16 × 54

= 0. 2 × 0. 1875 × 54

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The number of cuts would be 2

b. To determine the number of pieces of 20' angle to have 100 pieces we would divide the initial number of pieces by the angle

= 100/ 20

= 5 pieces

The number of pieces of 20' angle is 5

Thus, the number of cuts from a 20% length is 2 and the number of 20' angle are needed in order to have the 100 pieces is 5

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1 year ago
What is 3/8 + 1/8 - 2/7 + 1/4
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6 0
3 years ago
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